# Chapter 16: The Killing Form: A Powerful Insight into Global Lie Group Topology and Abstract Group Structure

The Killing Form is named after Wilhelm Killing (1847 – 1923), thus explaining its otherwise startling (for English speakers) name. Riemannian geometry even has Killing Fields, i.e. vector fields whose flows conserve the metric for the Riemannian manifold.

The Killing form is a homogeneous, billinear, symmetric binary real-valued form defined on a Lie algebra $\g$ and is the essence of simplicity; it is simply the following.

Definition 16.1 (Killing Form on a Lie Algebra):

Let $\g$ be a Lie algebra over a field $\mathbb{K}$. Then the Killing form on $\g$ is the billinear, binary $\mathbb{K}$-valued form:

and yet the Killing form sometimes yields deep insight into a Lie group’s global topology and abstract group structure. For example, as we shall see, a connected Lie group is compact if and only if its Killing form is negative definite, thus we see that some global topological information lies encoded in the Lie algebra, even though, as we have seen, we need to specify a Lie group’s fundamental group to fully understand its topology. Likewise, the Cartan criterion shows that a group is semisimple (Definition 15.8) if and only if the Killing form is nondegenerate as we shall see. In this case, we can further undertake a Cartan decomposition, splitting the Lie algebra $\g$ of Lie group $\G$ into two orthogonal subspaces $\g=\h\oplus\h^\perp$ such that $\h$ is a Lie algebra whereon the Killing form is negative definite (and thus the group $\H = \bigcup\limits_{k=1}^\infty \exp(\h)^k=\exp(\h)$ is compact) and the Killing from on $\h^\perp$ is positive definite. This decomposition generalises the polar decomposition of a matrix and indeed in a matrix group, the decomposition of $\gamma = \zeta\,\xi$ where $\zeta = e^X,\,X\in\h$ and $\xi=e^Y,\,Y\in\h^\perp$ is precisely the polar decomposition of $\gamma$. These notions can be most readily explored for $SO^+(1,\,3)$, the subject of Example 1.10.

The definition seems somewhat mysterious and arbitrary – even dippy – and it is hard to see the Killing form’s deep meanings from the bare, mechanical definition. So we first note an overview of all the concepts that we shall study in the following:

1. The Killing form is an invariant (also called associative) symmetric billinear form $\mathscr{K}:\g\times\g\to\R$ on a Lie algebra. That is, the Lie bracket “slides” within the Killing form through the “associative law” $\mathscr{K}([X,\,Y],\,Z) = \mathscr{K}(X,\,[Y,\,Z])=-\mathscr{K}(X,\,[Y,\,Z])$;
2. In a simple Lie algebra $\g$, Property 1. uniquely defines the Killing form, to within an arbitrary scaling constant;
3. In a semisimple Lie algebra $\g=\bigoplus\limits_{k=1}^M\,\g_k$ comprising $M$ simple direct sum summands, Property 1. defines the Killing form to within $M$ arbitrary scaling constants, each applied to the restriction of the invariant symmetric billinear form to each of the constituent summands;
4. A Lie group is semisimple if and only if its Killing form is nondegenerate; equivalently, if and only there exists a nondegenerate, invariant, symmetric billinear form (Cartan’s Criterion);
5. Every Abelian ideal of a Lie algebra $\g$ is always contained within the kernel of a degenerate Killing form;
6. The Killing form is “almost” a precise compactness detection machine: a Lie group with at most discrete center is compact if and only if its Killing form is negative definite. A compact group can have a negative semidefinite Killing form (as for the torus), and in this case the Killing form’s kernel is the same as the Lie algebra’s center so that the Lie algebra splits into the direct sum of the algebra’s center and an algebra with negative definite Killing form.

The trace being clearly fundamental to the Killing form, we first recall a few of the former’s properties

Lemma 16.3 (Properties of the Trace):

For the space $\mathcal{M}(N,\,\mathbb{K})$ of $N\times N$, $\mathbb{K}$-element square matrices (here $\mathbb{K}=\R,\,\mathbb{C}$), the trace $\operatorname{tr}:\mathcal{M}(N,\,\mathbb{K})\to\mathbb{K};\;\operatorname{tr}(X) = \sum\limits_{j=1}^N X_{j,\,j}$ has the following properties:

1. $\operatorname{tr}:\mathcal{M}(N,\,\mathbb{K})\to\mathbb{K}$ is $\mathbb{K}$-linear: $\operatorname{tr}(x\,X+y\,Y) = x\,\operatorname{tr}(X)+y\,\operatorname{tr}(Y);\;\forall\,X,\,Y\in\mathcal{M}(N,\,\mathbb{K}), x,\,y\in\mathbb{K}$;
2. The trace is the unique, to within an arbitrary scaling factor, Lie algebra homomorphism (respecting Lie brackets) of the form$\mathfrak{gl}(\V)\to\mathbb{K}$ where $\V$ is a vector space over a field $\mathbb{K}$ of characteristic nought;
3. The trace is the derivative at the identity of the Lie group homomorphism $\det:GL(\V)\to(\mathbb{K}\setminus\{0\},\,\times)$;
4. The trace of an $M$-fold product ($M\geq 2$) is invariant with respect to a cyclic permutation of the $M$ multiplicands, i.e. $\operatorname{tr}(X_1\,X_2\,\cdots\,X_{M-1}\,X_M) = \operatorname{tr}(X_2\,X_3\,\cdots\,X_M\,X_1)=\operatorname{tr}(X_M\,X_1\,\cdots\,X_{M-2}\,X_{M-1})$ for $\{X_j\}_{j=1}^M\subset\mathcal{M}(N,\,\mathbb{K})$;
5. The trace is invariant with respect to any similarity transformation i.e. if $Y = U\,X\,U^{-1}$ for any $X,\,Y\in\mathcal{M}(N,\,\mathbb{K})$ and invertible $U\in\mathcal{M}(N,\,\mathbb{K});\;\det U\neq0$ then $\operatorname{tr}(X)=\operatorname{tr}(Y)$;
6. $\operatorname{tr}(X)$ is the sum of all eigenvalues of $X$, each counted as many times as its multiplicity;
7. An $N\times N$ square matrix $X\in \mathcal{M}(N,\,\mathbb{K})$ is nilpotent if and only if the traces of the matrix and of all its strictly positive powers are nought, i.e. $\exists k\geq0\,\ni\,X^k=0\Leftrightarrow\operatorname{tr}(X^k)=0,\,\forall\,k\in\mathbb{N},k>0$ (by the Cayley-Hamilton theorem, this last condition is equivalent to $\operatorname{tr}(X^k)=0\,\forall\,k\in 1,\,2,\,\cdots\,N$);

Proof: Show Proof

For the proof of some of these, we use the fact that in $\mathcal{M}(N,\,\mathbb{C})$ any square matrix is unitarily similar to an upper triangular matrix. We shall prove this after the following.

1. Claim 1. is obvious;
2. A sketch of the proof of Claim 2. is as follows: we note that the image of a commutator $\rho([X,\,Y])$ under the homomorphism with the assumed property must be nought; now by finding commutators of matrices with one off-diagonal element each one can show that the image of such a matrix under the homomorphism must be nought to make the image all such commutators vanish. Thus a matrix with zeros along the diagonal is in $\ker(\rho)$ and $\rho$ must be a linear function of the leading diagonal elements alone. The weights in this function must also be equal if the image of a commutator of a diagonal matrix and an off diagonal matrix is to vanish always. Another proof is as follows: since $SL^+(N\,\mathbb{K})$ is simple, it must be inside the kernel of any homomorphism $\rho:GL^+(N\,\mathbb{K})\to (\mathbb{K},\,\times)$, because its image must have no normal subgroups other than the image itself or $\{1\}$. Therefore, write $\gamma\,\zeta\in GL(N\,\mathbb{K})$ uniquely as $\det(\gamma)\,\gamma_0\,\det(\zeta)\,\zeta_0$ where $\gamma_0,\,\zeta_0\in SL(N,\,\mathbb{K})$. Then $\rho(\gamma\,\zeta) = \rho(\det(\gamma)\,\det(\zeta)=\rho(\det(\gamma)\,\rho(\det(\zeta))$ and so we must find all $C^1$ solutions to the functional equation $f:\mathbb{K}\to\mathbb{K};\;f(u\,v) = f(u)\,f(v)$. For $\mathbb{K}=\R$ the most general solution is $f(u)=|u|^\lambda;\,\lambda\in\R$ and for $\mathbb{K}=\mathbb{C}$ we have $f(u)=u^\lambda;\,\lambda\in\mathbb{C}$. So the most general Lie group homomorphism to the multiplicative group is of the form $\gamma\mapsto(\det\gamma)^\lambda\,\lambda\in\mathbb{C},\,\mathbb{K}=\mathbb{C}$ or $\gamma\mapsto|\det\gamma|^\lambda\,\lambda\in\R,\,\mathbb{K}=\R$. Differentiating, the Lie map is therefore $X\mapsto\lambda\,\mathrm{tr}(X)$. Thus Claim 2. is proven;
3. Claim 3. is proven either by Jacobi’s formula (whose proof was sketched in Example Example 1.5) to infer $\det(e^{t\,X}) = \exp(t\,\mathrm{tr}(X))$ or alternatively as follows. $\det:GL(\V)\to(\R\setminus\{0\},\,*)$ is a Lie group homomorphism because the determinant is the volume of the hypercuboid defined by the rows of the matrix, therefore it respects multiplication. Therefore it induces its Lie map: a Lie algebra homomorphism $\mathfrak{gl}(\V)\to(\R,\,+)$. By Claim 2. there is only one such Lie map, to within a scaling constant;
4. Claim 4. follows from Claim 2: the image of a commutator is nought, therefore a cyclic permutation by moving the last multiplicand to the first position and shifting all others along must induce no change in $\rho$;
5. Claim 5. follows from Claim 4. on cyclically permuting the proposed similarity product $S\,A\,S^{-1}$ within the trace, so $\mathrm{tr}(S\,A\,S^{-1})= \mathrm{tr}(S^{-1}\,S\,A)=\mathrm{tr}(A)$. Alternatively, witness that the Lie group inner automorphism $GL(N,\mathbb{C})\to GL(N,\mathbb{C}); \zeta\mapsto\gamma\,\zeta\,\gamma^{-1},\,\gamma\in GL(N,\mathbb{C})$ is a Lie group inner automorphism with Lie map $\Ad(\gamma):\mathfrak{gl}(N,\,\mathbb{C})\to\mathfrak{gl}(N,\,\mathbb{C});\, X\mapsto\,\gamma\,X\,\gamma^{-1}$. Since $\det:GL(N,\mathbb{C})\to\R$ is a Lie group homomorphism, the cascaded Lie group homomorphism $\zeta\mapsto\gamma\,\zeta\,\gamma^{-1}\mapsto\det(\gamma\,\zeta\,\gamma^{-1})=\det(\zeta)$ (equal to $\zeta\mapsto\det\zeta$) has as its Lie map $\mathrm{tr}\circ\Ad(\gamma):\mathfrak{gl}(N,\,\mathbb{C})\to\R;\,\mathrm{tr}\circ\Ad(\gamma)X=\mathrm{tr}(\gamma\,X\,\gamma^{-1})$, which must be the same as the Lie map of $\zeta\mapsto\det\zeta$, namely $X\mapsto\mathrm{tr}(X)$ (by Claim 3.). Hence $\mathrm{tr}(X)=\mathrm{tr}(\gamma\,X\,\gamma^{-1})$;
6. If the base field is $\mathbb{C}$, then note that any square matrix is unitarily similar to an upper triangular matrix (proven below) and the eigenvalues must be on the leading diagonal of this matrix. Now apply Claim 5. Alternatively, use the Jordan normal form and reason likewise. If the base field is $\R$ then this claim must still hold as the sum of the eigenvalues has an imaginary part of nought, since the eigenvalues are either real or arise in conjugate pairs;
7. f the base field is $\mathbb{C}$, then note that any square matrix is unitarily similar to an upper triangular matrix. Therefore, a square matrix is nilpotent if and only if the leading diagonal elements of this upper triangular matrix are all nought. This is logically equivalent to the sum of all powers of the leading diagonal elements being nought. Now if the base field is $\R$, the sums are all real since the terms are either real or are in conjugate pairs.

So now we must simply prove that when $\mathbb{K}=\mathbb{C}$, any square matrix is unitarily similar to an upper (or lower) triangular matrix. A matrix $\mathbf{A}$ has at least one eigenvector $X$ – let’s make it a unit vector $\hat{X}$ – with eigenvalue $\lambda$ and, by the Gram-Schmidt procedure, build an orthonormal matrix of the form $\mathbf{P}_X=[\hat{X}\,\hat{Y}_2\,\cdots\,\hat{Y}_N]$ with columns $\hat{X}$,$\hat{Y}_2\,\cdots\,\hat{Y}_N$. Now work out $\mathbf{P}_X^{-1}\,\mathbf{A}\,\mathbf{P}_X=\mathbf{P}_X^\dagger\,\mathbf{A}\,\mathbf{P}_X$ and you find it is a matrix of the form:

$$\left(\begin{array}{c|c}\lambda & \cdots\\\hline \mathbf{0}&\mathbf{A}_1\end{array}\right)$$

where $\mathbf{A}_1$ is an $N-1\times N-1$ matrix. So now repeat the procedure with the matrix $\mathbf{A}_1$ and iterate, each iteration leaving a lower right square matrix of dimension one fewer than the one from the former iteration. When we reach the lower right corner, we have an upper triangular matrix, congruent to the original by the product of all the unitary transformations at each step. The proof for the lower triangular case is analogous; we simply use matrices of eigenvectors written as rows rather than columns.$\quad\square$

Now we recall the following properties of billinear forms; the Killing form clearly (by Claim 1. in Lemma 16.3) is symmetric in its arguments. A homogeneous billinear form $\mathscr{F}:\g\times\g\to\mathbb{K}$ (here $\mathbb{K} = \R,\,\mathbb{C}$, as appropriate) defined on a finite ($N$) dimensional vector space $\g$ over the field of scalars $\mathbb{K}$ can always be written in the form $\mathscr{F}(X,\,Y)=\left<X,\,f(Y)\right>$, by linearity in the first argument and the finite dimensional version of the Riesz representation theorem. By homogeneity and linearity in the second argument, $f(Y) = \mathbf{K}\, Y$, for some $N\times N$ matrix $\mathbf{K}$ of scalars, so that $\mathscr{F}(X,\,Y)=X^T \mathbf{K}\, Y$. If, further, the billinear form is Hermitian (symmetric if $\mathbb{K}=\R$) in its arguments (as is the Killing form) then $X^\dagger \mathbf{K}\, Y = Y^\dagger \mathbf{K}\, X = (Y^\dagger \mathbf{K}\, X)^\dagger = X^\dagger \mathbf{K}^\dagger\, Y,\;\forall\,X,\,Y\in\g$. Therefore, $\mathbf{K}^\dagger = \mathbf{K}$ and the finite dimensional $\mathbf{K}$ is orthogonally diagonalizable, with all real eigenvalues and eigenvectors corresponding to different eigenvalues are needfully orthogonal.

Let us recall the elementary proofs of these facts: specializing $\mathbb{K}=\R,\,\mathbb{C}$: if $X$ is an eigenvector of $\mathbf{K}$ with eigenvalue $\lambda$, then $\lambda\,X^\dagger\,X=X^\dagger\,\mathbf{K}\,X = (\mathbf{K}\,X)^\dagger\,X = \lambda^*\,X^\dagger\,X$, whence $\lambda\in\R$ (likewise, all eigenvalues of skew-Hermitian matrices are purely imaginary).

Let $\lambda_x,\,\lambda_y$ be different, real eigenvalues of the Hermitian $\mathbf{K}$, corresponding to eigenvectors $X,\,Y$, respectively. Then $\lambda_x\,X^\dagger\,Y =\lambda_x^*\,X^\dagger\,Y = (\mathbf{K}\,X)^\dagger \,Y=X^\dagger\,\mathbf{K}^\dagger\,Y = X^\dagger\,\mathbf{K}\,Y = \lambda_y\,X^\dagger\,Y$. Hence, we get $(\lambda_x-\lambda_y)\,X^\dagger\,Y = 0$, so that, if the eigenvalues are different, $X$ and $Y$ must be orthogonal. If the eigenvalues are the same, then any linear combination of $X$ and $Y$ is also an eigenvector, so that linear combinations of $X$ and $Y$ can be chosen to be orthogonal.

Now we write the characteristic equation for $\mathbf{K}$ and find that, by the fundamental theorem of algebra, there is at least one real eigenvalue $\lambda_0$ with eigenvector $X_0\in\g$; if the field $\mathbb{K}=\R$, the eigenvector is real too. Now let $Y$ be orthogonal to $X_0$; then $X_0^\dagger\,\mathbf{K}\,Y = (\mathbf{K}\,X_0)^\dagger\,Y=\lambda^* X_0^\dagger\,Y = 0$, so that $\mathbf{K}\,Y$ is also orthogonal to $X_0$. In other words, the $N-1$ dimensional orthogonal complement $\g_1=\{Y|\;X_0^\dagger\,Y=0\}\subset\g$ is an invariant subspace of $\mathbf{K}$ ($\mathbf{K}\,\g_1\subseteq\,\g_1$). We now find an orthonormal basis for $\g$ that includes $X_0$ as one of the basis vectors by the Gram-Schmidt process and impart the appropriate similarity transformation to express $\mathbf{K}$ relative to the new basis (comprising $X_0$ and the orthonormal basis for $\g_1$); this has the form $\mathbf{K}\mapsto\mathbf{U}\,\mathbf{K}\,\mathbf{U}^\dagger$ ($\mathbf{U}$ is real if the field $\mathbb{K}=\R$) and is clearly still Hermitian and, since the first basis member $X_0$ is an eigenvector, $\mathbf{K}$ is block-diagonal with the lower right $N-1\times N-1$ the Hermitian matrix for the restriction of $\mathbf{K}$ to the invariant subspace $\g_1$. So now we have our initial situation but in $N-1$-dimensions. We can clearly iterate this procedure, splitting off a new eigenvector $\hat{X}_m\in\g_{m-1}$ at the $m^{th}$ step, orthogonal to all the $\{\hat{X}_j\}_{j=1}^{m-1}$. So the process must resolve the whole of $\g$ into the span of $N$ orthogonal eigenvectors of $\mathbf{K}$. Clearly, if $\mathbf{K}$ is real and $\lambda$ a needfully real eigenvalue with corresponding eigenvector $X$, then $(\mathbf{K}-\lambda\,\id)\,X=0$ and also $(\mathbf{K}-\lambda\,\id)\,X^*=0$ so that both $\operatorname{Re}(X),\,\operatorname{Im}(X)$ are also eigenvectors. If $X$ and $X^*$ are linearly dependent, then any nonzero choice out of $\operatorname{Re}(X),\,\operatorname{Im}(X)$ will stand in the stead of $X$ and $X^*$; if $X$ and $X^*$ are linearly independent, then so must be $\operatorname{Re}(X),\,\operatorname{Im}(X)$ and so again we we can always choose linear combinations of $\operatorname{Re}(X),\,\operatorname{Im}(X)$ to be both orthogonal and stand in the stead of $X$ and $X^*$. So in the real case, which we have for the symmetric real Killing form when $\g$ is the Lie algebra of some Lie group $\G$, the eigenvectors of the matrix $\mathbf{K}$ of the Killing form are all real and orthogonal. Therefore:

Lemma 16.2 (Possibilities for the Killing Form):

Let $\g$ be an $N$-dimensional Lie algebra over the field $\mathbb{K}=\R,\,\mathbb{C}$ and $\mathscr{K}:\g\times\g\to\mathbb{K}$ the Killing form.

Then, for $\mathbb{K}=\R$:

\begin{equation}\label{KillingFormPossibilityLemma_1}\mathscr{K}(X,\,Y) = X^T\,\mathbf{K}\,Y = (\mathscr{O}\,X)^T \,\Lambda\,\mathscr{O}\,Y;\quad\mathbf{K}=\mathscr{O}^T\,\Lambda\,\mathscr{O}\end{equation}

where $\mathbf{K}\in\mathcal{M}(N,\,\R)$ is an $N\times N$, real element symmetric matrix ($\mathbf{K}=\mathbf{K}^T$), $\mathscr{O}$ is an $N\times N$, real element orthogonal matrix ($\mathscr{O}^T\,\mathscr{O} = \mathscr{O}\,\mathscr{O}^T = \id$) and $\Lambda=\operatorname{diag}(\lambda_1,\,\lambda_2,\,\cdots,\,\lambda_N)$ is a diagonal matrix of real eigenvalues.

For $\mathbb{K}=\mathbb{C}$:

\begin{equation}\label{KillingFormPossibilityLemma_2}\mathscr{K}(X,\,Y) = X^\dagger\,\mathbf{K}\,Y = (\mathscr{U}\,X)^\dagger \,\Lambda\,\mathscr{U}\,Y;\quad\mathbf{K}=\mathscr{U}^\dagger\,\Lambda\,\mathscr{U}\end{equation}

where $\mathbf{K}\in\mathcal{M}(N,\,\mathbb{C})$ is an $N\times N$, complex element, Hermitian matrix ($\mathbf{K}=\mathbf{K}^\dagger$), $\mathscr{U}$ is an $N\times N$, unitary matrix ($\mathscr{U}^\dagger\,\mathscr{U} = \mathscr{U}\,\mathscr{U}^\dagger = \id$) and $\Lambda=\operatorname{diag}(\lambda_1,\,\lambda_2,\,\cdots,\,\lambda_N)$ is a still diagonal matrix of real eigenvalues.

Thus for the Killing form, the following are mutually exclusive and exhaustive:

1. $\lambda_j<0;\,j\in1\cdots N$ and so $\mathscr{K}$ is negative definite;
2. The signs of $\lambda_j$ are mixed but none of the $\lambda_j$ are nought, $\mathscr{K}$ is thus nondegenerate and has a nontrivial signature and $\mathbf{K}$ is nonsingular;
3. The Killing form is degenerate, at least one of the $\lambda_j$ vanish and $\mathbf{K}$ is singular.

There are no Lie groups with positive definite Killing forms.$\quad\square$

The only fact in Lemma Lemma 16.2 above not yet proven by the discussion before the lemma is the fact there are no Lie groups with positive definite Killing form. The proof will come later.

In case 1. $-\mathscr{K}$ is a real inner product, a real inner product being a symmetric, billinear, positive definite function $\g\times\g\to\R$. In case 2. we don’t have an inner product: although there is no vector $X$ such that $\mathscr{K}(X,\,Y)=0\,\forall\,Y\in\g$, there are so-called (especially so-called in relativity theory) null vectors $X_1$ such that $\mathscr{K}(X_1,\,X_1)=0$ and so neither $\pm\mathscr{K}$ induce a norm.

Example 16.4 ($\mathfrak{so}(N)$ Killing Form):

As we have seen in Example Example 4.4, $SO(3)\cong\Ad(SO(3))$ and $\mathfrak{so}(3)\cong\ad(\mathfrak{so}(3))$. Let $\hat{S}_x = \ad(\hat{S}_x),\,\hat{S}_y= \ad(\hat{S}_y),\,\hat{S}_z= \ad(\hat{S}_z)\in\mathfrak{so}(3)\cong \ad(\mathfrak{so}(3))$ be a basis for $\mathfrak{so}(3)\cong \ad(\mathfrak{so}(3))$ as in Example Example 3.6; more generally, in $\mathfrak{so}(N)\cong \ad(\mathfrak{so}(N))$, choose the basis to be the $N\,(N-1)/2$ distinct matrices of the form:

\begin{equation}\label{SONKillingFormExample_1}S(\mu,\,\nu)_{j,\,k} = \left\{\begin{array}{llll}1&;&&j=\mu\,\text{and}\,k=\nu\\-1&;&&j=\nu\,\text{and}\,k=\mu\\0&;&&\text{otherwise}\end{array}\right.\end{equation}

By direct calculation, $\operatorname{tr}(S(\mu,\,\nu)\, S(s,\,\tau)) =\operatorname{tr}(\ad(S(\mu,\,\nu))\, \ad(S(s,\,\tau)))= -2\,\delta_{\mu\,s} \,\delta_{\nu\,\tau}$. Another quick way to see this, if the Frobenius matrix inner product and Frobenius matrix norm are wonted to you, is that $\operatorname{tr}(\ad(X)\,\ad(Y)) = -\operatorname{tr}(\ad(X)\,\ad(Y)^T);\;\forall\,X,\,Y\in \mathfrak{so}(N)$ so that the Killing form is the same as $-1$ times the matrix Frobenius inner product in the special case of the algebra $\mathfrak{so}(N)$ of skew-symmetric real matrices.

Thus, given $\eqref{SONKillingFormExample_1}$, the Killing form on $\mathfrak{so}(N)$ is simply $-1$ times the “obvious” or wonted, elementwise inner product between two matrices, i.e. $\mathscr{K}(X,\,Y) = \operatorname{vec}(X)^T\,\operatorname{vec}(Y)$ where $\operatorname{vec}:\mathcal{M}(N,\,\mathbb{K})\to\mathbb{K}^{N^2}$ means the operation that concatenates all the rows of a matrix into one, $N^2$ element row. So here, given the relationship with the Frobenius inner product of matrices, the Killing form bestows a Euclidean geometry on the Lie algebra. The group members act on the algebra by rigid, proper, Euclidean isometries.

## Invariance Properties of the Killing Form

Slightly unforeseen and unintuitive, given that the Lie bracket is not associative, is that the Killing form is a so-called invariant form (we’ll talk about where this name comes from after Lemma Lemma 16.6)i.e.:

Lemma 16.5 (Killing form is an Invariant Form):

Let $\g$ be a Lie algebra over the field $\mathbb{K}$ and $\mathscr{K}:\g\times\g\to\mathbb{K}$ its Killing form. Then $\mathbb{K}$ is an invariant form:

\begin{equation}\label{KillingFormInvariantFormLemma_1}\mathscr{K}([X,\,Y],\,Z) = \mathscr{K}(X,\,[Y,\,Z]);\;\forall\,X,\,Y,\,Z\in\g\end{equation}

Proof: Show Proof

Given the Jacobi identity (or a rewriting thereof, as we saw in Theorem Theorem 7.8) $\ad([X,\,Y]) = [\ad(X),\,\ad(Y)]$, we get $\mathscr{K}([X,\,Y],\,Z) = \operatorname{tr}(\ad(X)\,\ad(Y),\,\ad(Z) – \ad(Y)\,\ad(X),\,\ad(Z))$; now cyclically permute the multiplicands in the second term (as we can by property 2. in Lemma Lemma 16.3) so that $\mathscr{K}([X,\,Y],\,Z) = \operatorname{tr}(\ad(X)\,\ad(Y),\,\ad(Z) – \ad(X),\,\ad(Z)\,\ad(Y)) = \mathscr{K}(X,\,[Y,\,Z])$.$\quad\square$

Another simple key property is that if $\G$ is a connected Lie group and $\g$ its Lie algebra, then the Killing form on $\g$ is invariant when $\g$ undergoes the automorphism $\g\to\Ad(\gamma)\,\g$ for any $\gamma\in\G$. This is often tersely stated as “The Killing Form is $\Ad$-Invariant”.

Lemma 16.6 (Killing form is $\Ad$-Invariant):

Let $\G$ be a connected Lie group and $\g$ its Lie algebra. Then:

i.e. The Killing Form is $\Ad$-Invariant.

Proof: Show Proof

As shown in Example Example 7.24, $\Ad(\gamma):\g\to\g$ is a Lie bracket respecting automorphism and we apply $\eqref{TransformationOfLittleAdjointLaw}$ to show that its induced transformation law for $\ad(X)$ is $\ad(\Ad(\gamma)\,X) = \Ad(\gamma)\,\ad(X)\,\Ad(\gamma)^{-1}$. Then we apply this relationship as follows:

the last step following by property 5. of Lemma Lemma 16.3.$\quad\square$

A more direct and first principles proof is as follows:

Proof: Show Proof

We choose a suitable basis for $\g$ so that $\Ad(e^X)$ becomes the square matrix $\exp(\ad(X))$ for any $X\in\g$. Now, for any $X,\,Y,\,Z\in\g$ behold the Lie algebra member:

Now, on reckoning $\left.\partial_\tau\,Y_0(s,\,\tau)\right|_{\tau=0}$, line 3 and line 6 in $\eqref{KillingFormAdInvariantLemma_3}$ we show that:

and since this holds for all $Y\in\g$ we get:

So now, when $\Ad(e^{s\,Z})$ transforms $\g$, we see that the Killing form $\mathscr{K}(A,\,B)$ between $A,\,B\in\g$ transforms to (using $\eqref{KillingFormAdInvariantLemma_5}$ to get from the first to the second line):

the last line following by the invariance of the trace with respect to a similarity transformation (property 5. in Lemma Lemma 16.3). Lastly, since any $\gamma\in\G$ can be written as a finite product of the form $\gamma=\prod\limits_{k=1}^M\,e^{Z_k}$ for $Z_k\in\g$, and since $\Ad(\gamma) = \prod\limits_{k=1}^M\,\Ad(e^{Z_k}) = \prod\limits_{k=1}^M\,e^{\ad(Z_k)}$, we simply impart the transformation $\g\to\Ad(\gamma)\,\g$ in $M$ steps $\g\to\,e^{\ad(Z_M)}\,\g\to\,e^{\ad(Z_{M-1})}\,e^{\ad(Z_M)}\,\g\to\cdots \prod\limits_{k=1}^M\,e^{\ad(Z_k)}\g$ and apply the above reasoning at each step to show that $\mathscr{K}(\Ad(\gamma)\,A,\,\Ad(\gamma)\,B) = \mathscr{K}(A,\,B)$.$\quad\square$

The reason that the property in Lemma 16.5 is called “invariance” is that Lemma 16.5 is actually a “local” (“infinitessimal”) version of Lemma 16.6. Lemma 16.6 as $\mathscr{K}(\Ad(\gamma)\,X,\,\Ad(\gamma)\,Z) = \mathscr{K}(X,\,Z)$ for $X,\,Z\in\g$ and now we write $\gamma(\tau)=\exp(\tau\,Y)$. The relationship $\mathscr{K}(\Ad(\gamma)\,X,\,\Ad(\gamma)\,Z) = \mathscr{K}(X,\,Z)$ must hold for all $\tau\in\R,\,Y\in\g$ so, on equating the first order Taylor series terms (because $\mathscr{K}$ is billinear) we get the Lie algebraic statement that $\mathscr{K}([Y,\,X]\,\,Z)+\mathscr{K}(X,\,[Y,\,Z]) = 0$ which is Lemma 16.5. Conversely, if we begin with Lemma 16.5 in the form $\mathscr{K}(\ad(Y)\,X\,\,Z)+\mathscr{K}(X,\,\ad(Y)\,Z) = 0$ and form the universally convergent matrix Taylor series for $e^{\tau\,\ad(Y)}$ we derive Lemma 16.6 valid for any $e^X\in\G$ for $X\in\g$. Then, if $\mathscr{K}(\ad(Y)\,X\,\,Z)+\mathscr{K}(X,\,\ad(Y)\,Z) = 0 \Rightarrow \mathscr{K}(\Ad\left(e^{\tau\,\ad(Y)}\right)\,X,\,\Ad\left(e^{\tau\,\ad(Y)}\right)\,Z) = \mathscr{K}(X,\,Z)$ $\forall\,X,\,Y,\,Z\in\g$, we apply this again to find:

We repeat this process iteratively to show that $\mathscr{K}(\Ad(\gamma)\,X,\,\Ad(\gamma)\,Z) = \mathscr{K}(X,\,Z)$ is true for any $\gamma$ that is a finite product of terms of the form $e^{X_j}$ for $X_j\in\g$, i.e. true for all $\gamma$ in the identity connected component of a Lie group. We summarize:

Lemma 16.7 (Equivalence of Local and $\Ad$-Invariance):

Let $\G$ be a connected Lie group with Lie algebra $\g$. Then Lemma 16.5 and Lemma 16.6 are logically equivalent (can be derived from one-another).$\quad\square$

## Nondegenerate and Definite Killing Forms

Let’s look further at the property of nondegeneracy of the Killing form. In the light of Lemma Lemma 16.5, if the Killing form is degenerate, the Lie bracket bears the following striking likeness to the cross product in the following way. If $X,\,Y\in\g$ are vectors in a Lie algebra with such a Killing form, then write $\mathscr{K}(X,\,Y)$ as $\vec{X}\cdot\vec{Y}$ (even though the degenerate form may not be definite and therefore not an inner product) and write $[X,\,Y]$ as $\vec{X}\times \vec{Y}$. Then, from Lemma 16.5, we have $(\vec{X}\cdot \vec{Y})\times\vec{Z} = \vec{X}\cdot (\vec{Y}\times\vec{Z})$, which yields the wonted rule for the scalar triple product and is seen to generalize to Lie algebras with nondegenerate Killing forms. This in can be used to derive the formulas for the vector triple product in the special case of $\mathfrak{so}(3)$. In this case, $-\mathscr{K}$ is also a true inner product since $\mathscr{K}$ is negative definite. Nondegeneracy also implies some interesting things about the group’s center and its adjoint representation.

Lemma 16.8 (Nondegenerate Killing Form means Center is Discrete):

Let $\G$ be a Lie group with Lie algebra $\g$. Then the center $\mathscr{Z}(\G)$ is at most a discrete set if the Killing form on $\g$ is nondegenerate.

Proof: If the Killing form is nondegenerate, then $\forall\,X\in\g\,\exists\,Y\in\g\ni\mathrm{tr}(\ad(X),\,\ad(Y))\neq 0$. Therefore $\ad(X)\neq\Or$ and hence, by Theorem 7.19, the center$\mathscr{Z}(\G)$ is discrete.$\quad\square$

We can also impart a change of basis to a Lie algebra to rewrite the Killing form in a “canonical”, diagonal way. When a Lie algebra $\g$ is kitted with a basis that is “diagonalized” with respect to the Killing form, and when the latter is negatively or positively definite\footnote{We recall though we are yet to prove that a Lie algebra with positive definite Killing form is trivial}, it defines the wonted Euclidean distance given by the Pythagoras formula. We have already seen that the matrix $\mathbf{K}$ of the Killing form on a (real) Lie algebra $\g$ of a Lie group $\G$ is diagonalizable by an orthogonal matrix. To find the transformation law for $\mathbf{K}$ we use the following:

Lemma 16.9 (Billinear Form Transformation Law):

Let $\g$ be an $N$-dimensional Lie algebra over the field $\mathbb{K}$ ($\mathbb{K}=\R,\,\mathbb{C}$), $\mathscr{K}:\g\times\g\to\mathbb{K}$ a billinear form form and $\mathbb{K}$ the matrix of $\mathscr{K}$ with respect to some $\g$-basis $\{\hat{X}_j\}_{j=1}^N$. (i.e. $\mathbb{K}_{j\,k} = \mathrm{tr}(\ad(\hat{X}_j)\,\ad(\hat{X}_k))$. Suppose we make a change of basis defined by $\hat{X}_j = \sum\limits_{k=1}^N\,\rho_{j\,k}\,\hat{X}_k^\prime$ where $\rho$ is an invertible matrix and $\{\hat{X}_j^\prime\}_{j=1}^N$ the new $\g$-basis. Then, with respect to the new basis, the billinear form has the matrix:

\begin{equation}\label{BillinearFormTransformationLawLemma_1}\mathbf{K}^\prime=\rho^*\,\mathbf{K}^\prime\,\rho^T\end{equation}

Proof: Show Proof

This follows simply by writing choosing a $\g$-basis and writing the billinear form as a matrix equation: $\mathscr{K}(X,\,Y) = X^\dagger\,\mathbf{K}\,Y$ where now $X$ and $Y$ are column vectors containing the components, so if the $\g$-basis transforms following $\hat{X}_j = \sum\limits_{k=1}^N\,\rho_{j\,k}\,\hat{X}_k^\prime$, the components transform following $X^\prime = \rho^T\,X$. The transformed billinear form must then be such that $X^\dagger\,\mathbf{K}\,Y = {X^\prime}^\dagger\,\mathbf{K}^\prime\,Y^\prime$, so, on substituting the component transformations we get $X\,\rho^*\,\mathbf{K}^\prime\,\rho^T\,Y=X^\dagger\,\mathbf{K}\,Y;\;\forall\,X,\,Y\in\mathbb{K}^N$, whence the claimed transformation law follows.$\quad\square$

Take heed: Although it is tempting to think of the co-ordinate transformation $\rho$ as an endomorphism (which, of vector spaces, it is), $\rho$ in general does not respect Lie brackets. Indeed, when applied to the Killing form the whole point here is that we can choose a co-ordinate transformation that can alter the structure co-efficients to make the Killing form diagonal. Given that $\mathbf{K}$ is orthogonally (if the field$\mathbb{K}=\R$) or unitarily (if $\mathbb{K}=\mathbb{C}$) diagonalizable, we can write it as $\mathbf{K}={\rho^*}^\dagger\,\Lambda\,\rho^*$ for some orthonormal $\rho^*$. We then make the basis transformation $\hat{X}_j = \sum\limits_{k=1}^N\,\rho_{j\,k}^*\,\hat{X}_k^\prime$, whence, by Lemma 16.9, the Killing form now has the diagonal matrix $\Lambda$.

Now if we restrict ourselves to $\mathbb{K}=\R$, we can impart a further diagonal change of $\g$-basis defined by the matrix $\rho=\mathrm{diag}(\rho_1,\,\rho_2,\,\cdots)$ and then the matrix of the Killing form becomes $\rho\,\Lambda\,\rho^T = \mathrm{diag}(\rho_1^2\,\lambda_1\,\rho_2^2\,\lambda_2,\,\cdots)$. If $\lambda_j\neq0$ we can choose $\rho_j=\sqrt{|\lambda_j|^{-1}}$, so that the matrix of the Killing form becomes diagonal with only entries of $0$ or $\pm1$. We can then impart a change of $\g$-basis that re-orders the diagonal elements so that we group all the like eigenvalues together so that we have a diagonal Killing form matrix with only the values $0$, $-1$ or $+1$ along the leading diagonal. The number of $+1$, $-1$ and $0$ entries along the diagonal is uniquely defined when $\mathbb{K}=\R$ and equal to the number of positive, negative and nought eigenvalues of the matrix $\mathbf{K}$ defining a symmetric billinear form. This statement can be made even stronger, for we have only here thought about the diagonalization of the Killing form with real, orthogonal transformations. There could be other invertible, but nonorthogonal, transformations that diagonalize the Killing form when imparted by the transformation law of Lemma 16.9. Even then, the numbers of $\pm1$ and nought entries is an invariant of the billinear form. We cite without proof the well-kenned following:

Theorem 16.10 (Sylvester’s Law of Inertia):

Let $\mathscr{B}:\g\times\g\to\mathbb{K}$ be a symmetric billinear form on the linear space $\g$ over the field $\R$. Then there exist invertible co-ordinate transformations that diagonalize $\mathbf{K}$ through the transformation law of Lemma 16.9 so that $\mathbf{K}$ is a diagonal matrix with only entries $\pm1$ or $0$, and the numbers of positive, negative and zero entries is always the same and independent of which co-ordinate transformation does the diagonalization.

In particular, if the Killing form is nondegenerate, all the diagonal elements must be $\pm1$ and we have:

Lemma 16.11 (Nondegenerate Killing Form Implies $\G$ a Cover of $O^+(p,\,q)$):

An $N$-dimensional connected Lie group $\G$ with Lie algebra $\g$ whereon the Killing form is nondegenerate is a subgroup of a cover of $O^+(p,\,q)=SO(p,\,q)$ where $N=p+q$, i.e. of the identity-connected component of the matrix group of transformations that conserves the signatured billinear form with $p$ positive entries and $q$ negative entries.

Proof: Show Proof

The Killing form being nondegenerate, $\mathscr{Z}(\G) = \ker(\Ad)$ is at most discrete, so $\ker(\ad)=\{\Or\}$ so $\G$ is a discrete cover of $\Ad(\G)$ (Theorem 7.1 ). Now, as discussed above, we choose a $\g$-basis $\{\hat{X}_j\}_{j=0}^N$ that yields the nondegenerate Killing form in its “canonical” signatured form $\mathscr{K}(X,\,Y) = x_1\,y_1+ x_2\,y_2+\cdots+x_p\,y_p – x_{p+1}\,y_{p+1} – x_{p+2}\,y_{p+2}-\cdots-x_N\,y_N$, where the $x_j, \,y_j$ are the superposition weights of $X,\,Y$ with respect to the $\g$-basis. Since (Lemma 16.6) the Killing form is $\Ad$-invariant, when $\Ad(\gamma),\,\gamma\in\G$ acts on $\g$, it preserves the Killing form, i.e. the pseudo (signatured nondegenerate) inner product we have just defined between any two vectors $X,\,Y\in\g$, that is, $\Ad(\gamma)\in O(n,\,m)$ by definition and $\Ad(\G)\subseteq O(n,\,m)$. Thus $\G$ is a cover of $O(n,\,m)$ with at most a discrete fiber.$\quad\square$

## The Killing Form as a Compactness Detection Machine

We shall now prove that if the Killing form on the Lie algebra of a connected Lie group is negative definite, the Lie group is compact (“negative definite” also implies nondegenerate, because the eigenvalues of the matrix $\mathbf{K}$ of the Killing form must then be all strictly less than nought). This is a highly interesting fact: somehow the Lie algebra, which we have seen wholly encodes the group’s local nature, also encodes at least something about the global topology, even though, as we have seen, we do need to know the Lie group’s fundamental group to fully understand the global topology.

Theorem 16.12 (Definite Killing Form Implies Compactness):

Let $\G$ be a Lie group and $\mathscr{K}:\g\times \g\to\R$ the Killing form on the group’s Lie algebra $\g$. Suppose that $\mathscr{K}$ is definite (i.e. either $K(X,\,Y)\leq 0\,\forall\,X,\,Y\in\g$ or $K(X,\,Y)\geq 0\,\forall\,X,\,Y\in\g$ and $K(X,\,X)= 0 \,\Rightarrow\,X=\Or$ in either case). Then $\G$ is compact.

Proof: Show Proof

Firstly, if the Killing form is definite, then $\mathrm{tr}(\ad(X),\,\ad(X)) \neq 0;\;\forall\,X\in\g$, hence $\ad(X)\neq 0\,\forall,x\in\g$. Therefore (Theorem 7.1) $e^{\tau X}\in\G$ is noncentral for $|\tau|<\epsilon$ for some $\epsilon > 0$ and the group’s center can at most be discrete. So the identity connected component of $\G$ is at most a countable cover of $\Ad(\G)$, and we now study this latter group and show it is compact. After we have proven f $\Ad(\G)$ is compact, we then prove that $\G$ is compact by proving that the center $\ker(\Ad)$ of $\G$ is finite.

The function $\d:\g\times\g\to\R;\,\d(X,\,Y)=\pm\mathscr{K}(X-Y,\,X-Y)$ (-sign for negative definite, + for positive definite) is subadditive and defines a metric on $\g$ and $\langle\cdot,\cdot\rangle:\g\times\g\to\R;\,\langle X,\,Y\rangle=\pm\mathscr{K}(X,\,Y)$ defines a Euclidean inner product on $\g$. Also, reasoning as in Lemma 16.11, we see that $\Ad(\gamma),\,\gamma\in\G$ preserves this Euclidean inner product so that $\Ad(\G_\id)\subseteq O^+(N)=SO(N)$, the identity connected component of $O(N)$ when $\G_\id$ is the identity connected component of $\G$. We take as a known fact that $SO(N)$ is compact. Furthermore, $\Ad(\G_\id)\subseteq SO(N)$ is closed in the group topology of $SO(N)$ (as we shall show below in Lemma 16.13), so therefore $\Ad(\G_\id)$ is compact.

So now we consider the universal cover $\widetilde{SO(N)}$ of $SO(N)$. This is a finite, indeed double, cover. Since $\Ad(\G_\id)\subseteq SO(N)$, the universal cover $\widetilde{\Ad(\G_\id)}\subseteq \widetilde{SO(N)}$ is a subset of $\widetilde{SO(N)}$ so this must be a finite cover: each point of $\Ad(\G_\id)$ lifts to the same, finite, discrete (indeed two element) fiber that it does in $\tilde{SO(N)}$. Therefore the identity connected component $\G_\id$ of $\G$ must be compact. Likewise, all the connected components of $\G$ must be contained in the universal cover $\tilde{O(N)}$ of $O(N)$. $O(N)$ is a finite collection of connected components and its cover is also known to be finite. Therefore $\G$ is compact. $\quad\square$

So it remains to show that $\Ad(\G_\id)$ is a closed subgroup of $SO(N)$ when $\G$ has at most a discrete center.

Lemma 16.13 (Image under $\Ad$ of Discrete Centered Connected $\G$ is Closed in $SO(N)$):

Let $\G$ be an $N$-dimensional connected Lie group with at most a discrete center. Then $\Ad(\G)$ is a closed subgroup of $SO(N)$.

Proof: Show Proof

Because the exponential map $\exp:\mathfrak{so}(N)\to SO(N)$ is surjective[footnote 1], we can tell whether $\Ad(\G)$ is closed in $SO(N)$ by asking whether its inverse image $\mathcal{G}=\exp^{-1}(\Ad(\G)) = \{X|\,e^X \in\Ad(\G)\}$ under the exponential map is closed in the corresponding inverse image of $SO(N)$, namely $\mathfrak{so}(N)$. In particular, we must take heed that the intersection of an arbitrarily small open ball neighborhood of $\Or$, with respect to the group topology in $SO(N)$, is possibly a disconnected union of copies of the open ball $\{X\in\ad(\g)| \;\left\|X\right\|<\epsilon\}$ neighborhood of $\Or$ , with respect to the group topology in $\Ad(\G)$ (see, for comparison, the situation for the irrational slope torus subgroup in Figure 12.2). As with the torus example, the close stacking (in the topology of $SON(N)$) of copies of the ball open in the group topology of $\Ad(\G)$ arises when $\ad(\g)$ passes near (in the sense defined by the group topology of $\mathfrak{so}(N)$) to matrices in $\mathfrak{so}(N)\setminus\ad(\g)$ which are of the form $\mathscr{U}\,\mathrm{diag}(2\,\pi\,i\,k_1,\,2\,\pi\,i\,k_1,\,\cdots)\mathscr{U}^\dagger$ for $k_j\in\mathbb{Z}$.

To investigate this further, we split the Lie algebra $\mathfrak{so}(N)$ of $SO(N)\supseteq \Ad(\G)$ into the direct sum $\ad(\g)\oplus\ad(\g)^\perp$, where $\ad(\g)^\perp$ is the orthogonal complement defined by the inner product induced by the known negative definite Killing form $\mathscr{K}_{SO(N)}$ for the group $SO(N)$. Now, consider any $X_0\in\ad(\g)$ and we seek any $Y\in\ad(\g)^\perp$ such that $e^{X_0+Y}\in \Ad(\G)$. We consider the $C^\omega$ path through $\mathcal{G}$ defined by:

The map $\Ad(e^{\tau\,Z})$ is a continuous map on $\mathcal{G}$, indeed an isometry of $\mathcal{G}$. Therefore, the tangent at the identity belongs to $\ad(\g)$, hence $[Z,\,X_0+Y]\in\ad(\g)\,\forall\,Z\in\ad(\g)$. Since $X_0\in\ad(\g)$ we see that $[Z,\,Y]\in\ad(\g),\,\forall\,Z\in\ad(\g)$. But now $Y\in\ad(\g)^\perp$ so that $\mathscr{K}_{SO(N)}(Y,\,\ad(\g))=\{0\}$ and then Lemma 16.5 shows that:

whence $[Z,\,Y]\in\ad(\g)^\perp,\,\forall\,Z\in\ad(\g)$ and so $[Z,\,Y]\in\ad(\g)^\perp\bigcap\ad(\g)=\{\Or\},\,\forall\,Z\in\ad(\g)$. In other words, $Y\not\in\ad(\g)$ commutes with all of $\ad(\g)$ so that, if $e^{X_0+Y}=e^{X_0}\,e^{Y}=e^{Y}\,e^{X_0}$ and, since $e^{X_0+Y}\in \Ad(\g)$, we see that $e^{-X_0}\,e^{X_0+Y}=e^Y\in\Ad(\G)$ and, moreover, $e^Y$ commutes with all members of $\Ad(\G)$. In other words, $e^Y\in\mathscr{Z}(\Ad(\G))$ is central in $\Ad(\G)$ even though $Y\not\in\ad(\g)$. But now, since $\G$ is connected, by Lemma 7.25, $\Ad:\Ad(\g)\to \Ad(\Ad(\g))$ is a group isomorphism and indeed it is the identity map. So its kernel is trivial and since this kernel is the center of $\Ad(\G)$, we see that the center of $\Ad(\G)$ must be trivial. Therefore, we must have $e^Y=\id$ for any $Y\in\ad(\g)^\perp$ which is such that $e^{X+Y}\in\Ad(\G)$ for $X\in\ad(\g)$.

Thus we see that $\mathcal{G}$ is a stack of parallel (in the sense defined by the definite Killing form inner product) copies of $\ad(\g)$ and any spacing vector $Y\in\ad(\g)^\perp$ is such that $e^Y=\id$. So now we check whether the copies can be densely stacked. But $Y\in\mathscr{so}(N)$ must be of the form $2\,\pi\,i\,\mathscr{U}\,\mathrm{diag}(k_1,\,k_2,\,\cdots)\,\mathscr{U}^\dagger$ with $k_j\in\mathbb{Z}$ and $\mathscr{U}$ unitary if $e^Y=\id$ is to be true: if any ray $\{\tau\,\hat{Z}|\,\tau\in\R\}$ defined by a unit vector $\hat{Z}\in\ad(\g)^\perp$ contains a solution $Y_{\hat{Z}}$ such that $\exp\left(Y_{\hat{Z}}\right)=\id$, then the solutions are {\it periodic} along the ray with a minimum spacing (measured by the norm induced by the negative definite Killing form $\mathscr{K}_{SO(N)}$) of the form $2\,\pi\,\sqrt{k_1(\hat{Z})^2+k_2(\hat{Z})^2+\cdots}$ where the $k_j(\hat{Z})$ are integers greater than or equal to 1. Therefore, for every unit vector $\hat{Z}\in\ad(\g)^\perp$ there is a minimum spacing $\epsilon_{\hat{Z}}>0$ such that $\{\tau\,\hat{Z}|\,|\tau|<\epsilon_{\hat{Z}}\}$ is free of vectors $Y$ for which $e^Y=\id$ and this minimum spacing is greater than or equal to $2\,\pi$. Therefore, any neighborhood of $\Or$ in the group topology of $SO(N)$ contains a closed ball of the form $\bar{\mathcal{B}}_{\mathfrak{so}(N)}(\epsilon) = \{X\in\mathfrak{so}(N)|\,\left\|X\right\|\leq\epsilon\}$ where we choose $\epsilon<2\,\pi$ and this latter ball is both closed in the group topology for $SO(N)$ and also its only intersection with $\mathcal{G}$ is the ball $\bar{\mathcal{B}}_{\ad(\g)}(\epsilon) = \{X\in\ad(\g)|\,\left\|X\right\|\leq\epsilon\}$ which is in the group topology for $\Ad(\G)$. So the only limit points of $\bar{\mathcal{B}}_{\ad(\g)}(\epsilon)$ in the group topology for $SO(N)$ are also limit points in the group topology for $\ad(\g)$; since it is closed in the latter it must contain these limit points. So therefore $\mathcal{G}$ contains all its limit points and is thus closed in $\mathfrak{so}(N)$.$\quad\square$

Foonote: Any orthogonal matrix $\zeta$ is unitarily diagonalizable $\zeta=\mathscr{U}\Lambda\mathscr{U}^\dagger$ with $\mathscr{U}\in U(N)$ and $\Lambda$ containing eigenvalues on the unit circle, indeed the eigenvalues are either $\pm1$ or arise in complex conjugate pairs (since the characteristic equation of $\zeta$ has real co-efficients). Therefore, $\exp:\mathfrak{so}(N)\to SO(N)$ is surjective: $\forall \zeta\in SO(N)\exists X\in\mathfrak{so}(N)\ni e^X=\zeta$: take $X = \mathscr{U}\log\Lambda\mathscr{U}^\dagger$ for any branch of the logarithm, taking heed that $\log\lambda_j$ exists for all the unit magnitude eigenvalues $\lambda_j$.

It is worth noting the differences between the present case and the noncompact irrational slope subgroups of the torus. In the torus we can of course decompose Lie algebra vectors into the form $X+Y$ where $X$ is in the Lie algebra for the irrational slope subgroup and $Y$ is “orthogonal” to $X$ (although not in the sense of a Killing from, which is identically nought for the torus) as we did above. But the fact that $e^Y$ is central now does not help us at all: this property doesn’t narrow the possibilities down a whit since all the group is Abelian, so all elements are central! This therefore is shown to be the key difference from the above.

## What’s So Special about the Killing form?

The Killing form’s simplicity bears with it a certain measure of mystery: why is such a simple, yet seemingly arbitrarily defined billinearform so meaningful? How does such a “blunt” (in the sense of having a big and highly nontrivial kernel) functional as the trace draw fine meaning from the Lie algebra in the way the Killing form seems to? The foregoing section begins to yield insight here: if one sifts carefully through the proof of Theorem 16.12, one sees that all its steps would work with any symmetric definite billinear form which is invariant in the sense of Lemma 16.5. Most of the Killing form’s useful properties are like this: given that the Killing form is invariant (Lemma 16.5), the question of whether or not the Killing form is degenerate and or definite becomes highly meaningful. So it is natural to ask how unique the Killing form is. The first result answering this question is that for a simple Lie group, the only symmetric, nondegenerate, invariant billinear form is a constant times the Killing form. To prove this result, we shall first need the famous Schur’s lemma on group representations. We have of course already met representation in the Adjoint representation of a Lie group (Definition 4.2) and the notion is general:

Definition 16.14 (Group Representation):

Let $G$ be a group (finite or infinite, Lie or otherwise) and $V$ a linear space over the field $\mathbb{K}$. Let $\rho:G\to GL(V)$ be a homomorphism from $G$ to the group of invertible transformations (automorphisms) of $V$. Then the triple $(G,\,\rho,\,V)$ is called a representation or a $G$-module of the group $G$. The dimension $\dim V$ is called the degree of the representation. we say $G$ acts on $V$ through $\rho$ and, if the homomorphism $rho$ is clear from the context, one often writes $\gamma\,X\in\V$ as a shorthand for the image of $X\in V$ under the action of $\gamma\in G$ through $\rho$, i.e. $\gamma\,X \stackrel{def}{=} \rho(\gamma)\,X$.

With representations, the notion of irreducible representation is especially important:

Definition 16.15 (Irreducible Representation):

A representation $(G,\,\rho,\,V)$ of $G$ is called irreducible if there is no nontrivial proper vector subspace $W\subset V$ that is left invariant by $G$, i.e. $(G,\,\rho,\,V)$ is irreducible if whenever$W\subset V$ is a vector subspace of $V$ such that $G,W = \{\rho(\gamma)\,X|\,\gamma\in G;\,X\in W\}= W$, then either $W=\{\Or\}$ or $W=V$.

If a representation of $G$ is not irreducible, then we have the following notion:

Definition 16.17 (Subrepresentation):

Let $(G,\,\rho,\,V)$ be a representation of $G$ and suppose there is a vector subspace $G,W = \{\rho(\gamma)\,X|\,\gamma\in G;\,X\in W\}= W$. Then $(G,\,\rho,\,W)$ is called a subrepresentation of $(G,\,\rho,\,V)$.

so that we can define an irreducible representation in perhaps wieldier terms than those of Definition 16.15 as follows:

Definition 16.17 (Irreducible Representation, version 2):

A representation $(G,\,\rho,\,V)$ of $G$ is called irreducible if its only subrepresentations are itself and the trivial representation $(G,\,\rho,\,\{\Or\})$.

Also, suppose for a particular group $G$ we have two representations $(G,\,\rho,\,V)$ and $(G,\,\phi,\,W)$ for two different vector spaces $V$ and $W$ over the same field $\mathbb{K}$. Naturally, there are many homogeneous linear maps between the two spaces $V$ and $W$, but there are ones that have special significance for a the particular group $G$.

Definition 16.18 ($G$-Linear Map, Intertwining Map):

Let $(G,\,\rho,\,V)$ and $(G,\,\phi,\,W)$ be two representations of the group $G$ for two, in-general-different, vector spaces $V$ and $W$ over the same field $\mathbb{K}$. Then a map $\psi:V\to W$ is called $G$-linear if the diagram below:

\begin{equation}\label{IntertwiningMapDefinitionGraph}\require{AMScd}\begin{CD}V @>{\psi}>> W\\@V\rho(\gamma)VV @VV\phi(\gamma)V \\V @>{\psi}>> W\end{CD}\end{equation}

commutes for all $\gamma\in G$. Otherwise put, $\phi(\gamma)\circ\psi:V\to W$ is precisely the same map as $\psi\circ\rho(\gamma):V\to W$ for every $\gamma\in G$. Other names for such a map are intertwining map or $G$-module homomorphism.

With these definitions made, we are ready to state and prove the famous Schur’s lemma; we shall give two versions:

Theorem 16.19 (Schur’s Lemma):

Let $(G,\,\rho,\,V)$ and $(G,\,\phi,\,W)$ be two irreducible representations of the group $G$ for two, in-general-different, vector spaces $V$ and $W$ over the same field $\mathbb{K}$ and let further $\psi:V\to W$ be an intertwining map as in Definition 16.18.

1. Then $\psi$ must be either the trivial endomorphism ($\psi(X)=\Or,\,\forall X\in V$) or $\psi$ is bijective thus invertible;
2. If further the field $\mathbb{K}$ is algebraically closed (e.g. $\mathbb{K}=\mathbb{C}$) then $\psi:V\to W$ is proportional to the identity map, i.e. a pure scaling map;
3. Let $G\subseteq GL(V)$ be any group of automorphisms of the vector space $V$ over the algebraically closed field $\mathbb{K}$ (i.e. equivalent to a group of matrices with elements in $\mathbb{K}$), such that there is no proper linear subspace $W\subset V$ which is left invariant by all the members of $G$ (the smallest invariant subspace of the matrix group members is the whole of $V$). Let $P$ be a matrix (not needfully invertible) that commutes with all of $G$. Then $P$ is proportional to the identity matrix.

Proof: Show Proof

To Prove Claim 1.: Consider $X\in\ker(\psi)\subseteq V$ and $\rho(\gamma)\in GL(V)$ for any $\gamma\in G$. Since $\psi$ is an intertwining map, $\psi\circ \rho(\gamma) X = \phi(\gamma)\circ\psi(X) =\Or$ so that $\rho(\gamma) \,X\in\ker(\psi)\,\forall\,\gamma\in G$, thus $G\,\ker(\psi) = \ker(\psi)$ and thus $(G,\,\rho,\,\ker(\psi))$ is a subrepresentation of $(G,\,\rho,\,V)$. Therefore, by assumption of irreducibility, $\ker(\psi)=V$ or $\ker(\psi)=\{\Or\}$. Likewise, consider $Y\in\mathrm{image}(\psi)\subseteq W$ and $\phi(\gamma)\in GL(W)$ for any $\gamma\in G$. The $\exists Z\in V \ni \psi(Z)=Y$. Now, intertwining assumption, $\phi(\gamma)\circ\psi(Z) = \psi\circ \rho(\gamma) Z \in \mathrm{image}(\psi)$ by definition of the image, i.e. $G\,\mathrm{image}(\psi) = G$ (here through the group action induced by $\phi$), $(G,\,\phi,\,\mathrm{image}(\psi))$ is a subrepresentation of $(G,\,\phi,\,W)$ and therefore, by assumption of irreducibility, $\mathrm{image}(\psi)=W$ or $\mathrm{image}(\psi)=\{\Or\}$.

So we are left with four possibilities arising from the logical statement $\left(\ker(\psi)=V\text{ or }\ker(\psi)=\{\Or\}\right)\text{ and }\left(\mathrm{image}(\psi)=W\text{ or }\mathrm{image}(\psi)=\{\Or\}\right)$:

1. $\ker(\psi)=V$ and $\mathrm{image}(\psi)=\{\Or\}$ and $\psi$ is the zero endomorphism;
2. $\ker(\psi)=V$ and $\mathrm{image}(\psi)=W$: impossible, since if $\ker(\phi)=V$ then $\phi(V)=\{\Or\}$;
3. $\ker(\psi)=\{\Or\}$ and $\mathrm{image}(\psi) = W$ and thus $\psi$ is one-to-one and onto;
4. $\ker(\psi)=\{\Or\}$ and $\mathrm{image}(\psi)=\{\Or\}$: impossible since $\mathrm{image}(\psi)=\{\Or\}\Rightarrow \phi(V)=\{\Or\}$ and then $\ker(\psi)=V$.

and there are only two possibilities left, those of Claim (1).

To Prove Claim 2.: If $W\cong V$ and if $\mathbb{K}$ is algebraically closed, the square matrix $\boldsymbol{\Psi}$ of $\psi$ has at least one eigenvalue $\lambda\in\mathbb{K}$, since the polynomial characteristic equation must have a zero by assumption of algebraic closure. But $\boldsymbol{\Psi}-\lambda\,\id$ is the matrix of an endomorphism $V\to W$ that fulfills the conditions of Claim (1), so therefore either $\ker(\boldsymbol{\Psi}-\lambda\,\id)=\{\Or\}$ or $\ker(\boldsymbol{\Psi}-\lambda\,\id)=V$. The first possibility $\ker(\boldsymbol{\Psi}-\lambda\,\id)=\{\Or\}$ is only possible if $\lambda=0$, by definition of the eigenvalue. So therefore either $\boldsymbol{\Psi}$ is the zero matrix, or $\boldsymbol{\Psi}=\lambda\,\id$. This proves Claim (2).

To Prove Claim 3.: Here we put $W=V$, $\phi=\rho=\id$ to get our representation of the group $G$ of matrices (which belong to $GL(V)$). Since $P$ commutes with any $\gamma\in G$, $P$ is an intertwining map of Definition 16.18 so now $(G,\,\rho=\id,\,V)$, $(G,\,\phi=\id,\,W=V)$ and $\psi=P$ fulfill the conditions of Claims 1. and 2., therefore $P=\Or$ or $P=\lambda\,\id$, for some $\lambda\in\mathbb{K}$.$\quad\square$

At last we are ready to answer the question of how unqiue the Killing form is in its particular properties. For simple Lie algebras over $\R$:

Theorem 16.20 (Symmetric, Invariant Billinear Forms are Unique to Within a Scale on a Simple Lie Algebra):

Let $\g$ be an $N$-dimensional simple Lie algebra over an algebraically closed field $\mathbb{K}$. Then any symmetric, invariant (in the sense of Lemma 16.5) billinear form $\mathscr{B}:\g\times\g\to\mathbb{K}$ is unique to within a scaling constant. In particular, all such forms are proportional to the Killing form.

Proof: Show Proof

Firstly, a symmetric, invariant, billinear form on a simple Lie algebra cannot be degenerate, because a nontrivial kernel of such a form would be an ideal, gainsaying the assumption of simplehood. To understand this, let $X\in\ker(\mathscr{B})$ and let $Y,\,Z\in\g$. Then, by invariance $\mathscr{B}([X,\,Y],\,Z)=\mathscr{B}(X,\,[Y,\,Z])$ and this quantity is nought, because $X$ is in the kernel. Hence $[X,\,Y]=0\,\forall\,Y\in\g\Rightarrow\,X\in\ker(\mathscr{B})$.

Now let $\mathscr{B}$ and $\tilde{\mathscr{B}}$ be two symmetric, invariant, billinear forms on $\g$. Then we can write $\tilde{\mathscr{B}}(X,\,Y) = \mathscr{B}(X,\,\mathbf{P}\,Y)$, where $\mathbf{P}$ is an $N\times N$ matrix. To understand this, write the billinear forms in matrix notation $\tilde{\mathscr{B}}(X,\,Y) = X^T\,\tilde{\mathbf{B}}\,Y$, $\mathscr{B}(X,\,Y) = X^T\,\mathbf{B}\,Y$; since neither $\mathscr{B}$ nor $\tilde{\mathscr{B}}$ can be degenerate, we can simply put $\mathbf{P}=\tilde{\mathbf{B}}^{-1}\,\mathbf{B}$. Now let $X,\,Y,\,Z\in\g$ then:

and therefore $0=\tilde{\mathscr{B}}(X,\,\left(\ad(Z)-\mathbf{P}^{-1}\,\ad(Z)\,\mathbf{P}\right)\,Y)$ for all $X,\,Y\in\g$. Since $\tilde{\mathscr{B}}$ is nondegenerate, this implies the matrix equation $\ad(Z)-\mathbf{P}^{-1}\,\ad(Z)\,\mathbf{P}=0$, i.e. $\mathbf{P}$ commutes with $\ad(Z)$ for all $Z\in\g$, a statement that is true if and only if $\mathbf{P}$ commutes with $e^{\tau\,\ad(Z)}$ for every $Z\in\g$ and $\tau\in\mathbb{K}$. Now if there were some linear subspace $W\subset\g$ such that $e^{\tau\,\ad(Z)}\,W=W$ for all $Z\in\g$ and $\tau\in\mathbb{K}$ then let $Y\in W$ and we see that $e^{\tau\,\ad(Z)}\,Y\in W\,\forall\tau\in\mathbb{K}$ only if $\ad(Z),\,Y=[Z,\,Y]\in W$, that is, $W$ is an ideal of $\g$. So, by simplehood of $\g$, $W=\{\Or\}$ (impossible, since $e^{\tau\,\ad(Z)}$ is always invertible thus bijective) or $W=\g$. That is, behold the matrix group $\G=\{\exp(\ad(Z))|\,Z\in\g\}$: then $(\G,\,\id,\,\g)$ is an irreducible representation of $\G$. This means that $\mathbf{P}$ and $\G$ fulfill the conditions of Claim 3. in Schur’s Lemma (Theorem 16.19) and we must have $\mathbf{P}=\lambda\,\id$, for some scaling factor $\lambda\in\mathbb{K}$. Conversely, any scalar multiple of a symmetric, invariant, billinear form on $\g$ is symmetric, invariant, billinear form on $\g$.$\quad\square$

Now one can broaden the above to semisimple Lie algebras in a straightforward way.

Theorem 16.21 (Constraints on Invariant Forms in Semisimple Algebras):

Let $\g$ be a semisimple Lie algebra over an algebraically closed field $\mathbb{K}$, i.e. $\g=\g_1\oplus\g_2\oplus\g_3\oplus\cdots$ is a direct sum of simple Lie algebras. Suppose that a basis for $\g$ is ordered to group the basis members for each simple component together. Then:

1. The Killing form on $\g$ is nondegenerate;
2. The matrix $\mathbf{K}$ of the Killing form is block-diagonal in this basis and is of the form: \begin{equation}\label{SemiSimpleAlgebraKillingFormUniquenessTheorem_1}\mathbf{K} = \left(\begin{array}{c|c|c|c}\mathbf{K}_1&\Or&\Or&\cdots\\\hline\Or&\mathbf{K}_2&\Or&\cdots\\\hline\Or&\Or&\mathbf{K}_3&\cdots\\\hline\vdots&\vdots&\vdots&\vdots\end{array}\right)\end{equation} where the diagonal blocks are compatible with groupings of basis members for the simple direct summand Lie algebras $\g_j$ and $\mathbf{K}_j$ are the matrices of the Killing form restricted to the subalgebras $\g_j$;
3. Any symmetric, invariant (in the sense of Lemma 16.5) billinear form $\mathscr{B}:\g\times\g\to\mathbb{K}$ has a matrix that is of the form $\mathbf{P}\,\mathbf{K}$ where $\mathbf{P}$ is block diagonal in the way compatible with the blocks of $\mathbf{K}$ and of the form:\begin{equation}\label{SemiSimpleAlgebraKillingFormUniquenessTheorem_2}\mathbf{P} = \left(\begin{array}{c|c|c|c}\lambda_1\,\id_1&\Or&\Or&\cdots\\\hline\Or&\lambda_2\,\id_2&\Or&\cdots\\\hline\Or&\Or&\lambda_3\,\id_3&\cdots\\\hline\vdots&\vdots&\vdots&\vdots\end{array}\right)\end{equation} where the $\id_j$ is the $\dim\g_j\times\dim\g_j$ identity matrix.

The above is often more tersely stated: A symmetric, invariant billinear form on a semisimple Lie algebra comprising $M$ direct summands is unique to within $M$ separate scaling constants each applied to the diagonal blocks of the Killing form. Or: the restriction of an invariant, symmetric billinear form on a semisimple Lie algebra to each of the simple summands in the direct sum is unique to within a scaling constant, but one is free to choose the scaling constants for the summands independently – there are as many degrees of scaling freedom as there are direct sum summands.

Proof: Show Proof

{\bf To prove Claims 1. and 2.}: In the stated basis for $\g=\g_1\oplus\g_2\oplus\g_3\oplus\cdots$, the matrix of any $\ad(Z)$ for $Z\in\g$ can be written as a block-diagonal matrix:

\end{equation}

where $Z=(Z_1,\,Z_2,\,Z_3,\,\cdots)$ and the $Z_j$ are the projections of $Z$ onto the constituent, simple, direct summands $\g_1,\,\g_2,\,\g_3,\,\cdots$. The stated form of the Killing form follows and, since the $\ad(Z_j)$ come from simple Lie algebras, the diagonal blocks $\ad(Z_1)$ must be nonsingular by the nondegeneracy argument put forward in Theorem 16.20 applied to each of the $\g_j$.

{\bf To prove Claim 3.}: Arguing with $\g$ members that have only one nonzero projection into one of the summands $\g_j$ at a time as in Theorem 16.20 we can show that the generalized matrix $\mathbf{P}$ analogous to the one in Theorem 16.20 is also block-diagonal with blocks compatible with the matrix in $\eqref{BlockDiagonalAdjoint}$ and must be a matrix of the claimed form.$\quad\square$

We have thus seen that a semisimple algebra has a nondegenerate Killing form and, by Theorem 16.21, all symmetric, invariant billinear forms on a semisimple algebra are nondegenerate. The converse also holds, i.e. that if a Lie algebra has a nondegenerate, symmetric, invariant billinear form then it must be semisimple. Before proving this, we shall need two simple results:

Lemma 16.22 (Restriction of Killing Form to an Ideal):

Let $\g$ be a Lie algebra and $\h\subset\g$ an ideal. Then the Killing form on $\h$ is simply the restriction of the Killing form on $\g$ to $\h$.

Proof: Let $X,\, Y\in\h$. Since $\ad(X)\,\ad(Y):\g\to\h$, only rows corresponding to the subspace $\h$ can be nonzero in the matrix of $\ad(X)\,\ad(Y)$. So the trace of $\ad(X)\,\ad(Y)$ computed as an endomorphism of $\h$ alone is the same as that of $\ad(X)\,\ad(Y)$ computed as an endomorphism of $\g$.$\quad\square$

Lemma 16.23 (Abelian ideals contained withing Killing form’s kernel):

Let $\g$ be a Lie algebra $\g$ over the field $\mathbb{K}$ with Killing form $\mathscr{K}:\g\times \g\to\mathbb{K}$. Then any Abelian ideal $\h\subseteq\g$ of $\g$ is contained within the kernel of the Killing form $\ker(\mathscr{K}) = \{X| \mathscr{K}(X,\,Y)=0\,\forall\,Y\in\g\}$.

Proof: Show Proof

If $\h$ is an Abelian ideal then $[\h,\,\g]\subseteq\h$ (i.e. $\ad(X)\,Y\in\h\,\forall\,Y\in\g\,\forall\,X\in\h$) and $[\h,\,\h]=\{\Or\}$ (i.e. $\ad(X)\,Z=\Or,\,\forall X,\,Z\in\h$). So now put $U=\ad(Y)\,\ad(X)\,Z$ for any $X\in\h$ and $Y,\,Z\in\g$ so that $U\in\h$ and then $\ad(X)\,\ad(Y)\,\ad(X)\,Z=0,\,\forall\,X\in\h,\,\forall\,Y,\,Z\in\g$. That is, $\ad(X)\,\ad(Y)\,\ad(X)=\Or$ whence $(\ad(X)\,\ad(Y))^2=\Or$. Thus the matrix $\ad(X)\,\ad(Y)$ is nilpotent, so its trace must be nought (Lemma 16.3) therefore $\mathscr{K}(X,\,Y)=0$ for any $Y\in\g$ when $X$ belongs to an Abelian ideal. Therefore $X\in\ker(\mathscr{K})$ as claimed.$\quad\square$

The kernel of the Killing form (or of any billinear form) is sometimes called its radical. In particular, a Lie algebra with nondegenerate Killing form cannot contain any Abelian ideals. With this observation in mind, we look at the following elegant proof that a Lie algebra’s nondegeneracy implies its semisimplicity due to Jean Dieudonné (1906-1992)

Theorem 16.24 ((Dieudonné, 1953)\cite{Dieudonne} Nondegenerate Killing Form Implies Semisimple Lie Algebra):

Let $\g$ be a Lie algebra over a field $\mathbb{K}$ with a nondegenerate, symmetric, invariant billinear form $\mathscr{B}:\g\times\g\to\mathbb{K}$. Then $\g$ is semisimple, i.e. the direct sum of simple Lie algebras.

Proof: Show Proof

Let $\mathfrak{f}$ be a minimal ideal of $\g$, that is, one with no proper ideals (so $\mathfrak{f}$ is a simple Lie algebra). $[\mathfrak{f},\,\mathfrak{f}]\subseteq\mathfrak{f}$ is another ideal, but since $\mathfrak{f}$ was assumed minimal, we must then have $[\mathfrak{f},\,\mathfrak{f}]=\{\Or\}$ or $[\mathfrak{f},\,\mathfrak{f}]=\mathfrak{f}$. But our nondegeneracy assumption and the foregoing Lemma 16.23 rule out the possibility $[\mathfrak{f},\,\mathfrak{f}]=\{\Or\}$ whence we must have $[\mathfrak{f},\,\mathfrak{f}]=\mathfrak{f}$.

So now let us look at the linear subspace $\tilde{\mathfrak{f}}=\{X\in\g|\,\mathscr{K}(X,\,Y)=0\,\forall\,Y\in\mathfrak{f}\}$, the set of Lie algebra members “orthogonal” to $\mathfrak{f}$\footnote{Although recall that our billinear form has not been assumed definite, so we cannot assume that we can decompose $\g$ as $\g=\mathfrak{f}\oplus\tilde{\mathfrak{f}}$ as we can with a true inner product, even though it turns out to be true in the present case which the rest of the proof of course shows.}. Then, if $Z\in\tilde{\mathfrak{f}}$ and $X\in\mathfrak{f}$, we have $\mathscr{K}([Z,\,Y],\,X)=\mathscr{K}(Z,\,[Y,\,X])$ because the Killing form is invariant (Lemma 16.5) and therefore $\mathscr{K}([Z,\,Y],\,X)=\mathscr{K}(Z,\,[Y,\,X])=0$ because $[Y,\,X]\in\mathfrak{f}$ ($X\in\mathfrak{f}$ and $\mathfrak{f}$ an ideal) so therefore $[Z,\,Y]\in\tilde{\mathfrak{f}}$ for every $Y\in\g$. Therefore $\tilde{\mathfrak{f}}$ is an ideal also. The intersection of two ideals being again an ideal, we must have either:

1. $\mathfrak{f}\bigcap\tilde{\mathfrak{f}}=\{\Or\}$; or
2. $\mathfrak{f}\bigcap\tilde{\mathfrak{f}}=\mathfrak{f}$ because $\mathfrak{f}$ was assumed minimal.

However, alternative 2. can be shown impossible as follows. If $U,\,V\in\mathfrak{f}\subseteq\tilde{\mathfrak{f}}$ then, by definition of $\tilde{\mathfrak{f}}$, we have $\mathscr{B}(U,\,V)=0$. Now let $X\in\mathfrak{f}\subseteq\tilde{\mathfrak{f}}$ then $X=\sum\limits_k [Y_k,\,Z_k]$ for $Y_k,\,Z_k\in\mathfrak{f}$ because we have already seen that $[\mathfrak{f},\,\mathfrak{f}]=\mathfrak{f}$. So then for every $A\in\g$ we have $\mathscr{B}(X,\,A) = \sum\limits_k \mathscr{B}([Y_k,\,Z_k],\,A) = \sum\limits_k \mathscr{B}(Y_k,\,[Z_k,\,A])$ but $\mathfrak{f}$ is an ideal so $[Z_k,\,A]\in\mathfrak{f}$. We have already seen our assumption that $\mathfrak{f}\subseteq\tilde{\mathfrak{f}}$ implies $\mathscr{B}(U,\,V)=0\,\forall U,\,V\in\mathfrak{f}$, so we then have $\mathscr{B}(Y_k,\,[Z_k,\,A])=0$. Therefore $\mathscr{B}(X,\,A) =0\,\forall\,A\in\g$, gainsaying our assumption that $\mathscr{B}$ was degenerate. Hence of the two possibilities above, we must have $\mathfrak{f}\bigcap\tilde{\mathfrak{f}}=\{\Or\}$.

Now, since $\mathscr{B}$ is nondegenerate, we have $\g=\mathfrak{f}\oplus\tilde{\mathfrak{f}}$ and we have split $\g$ into the direct sum of a simple Lie algebra $\mathfrak{f}$ and its orthogonal complement. By this orthogonality, the matrix of $\mathscr{B}$ with respect to the union of a basis for $\mathfrak{f}$ and one for $\tilde{\mathfrak{f}}$ is block diagonal with blocks of dimensions $\dim\mathfrak{f}$ and $\dim\tilde{\mathfrak{f}}$ and so neither diagonal block can be a singular square matrix. Therefore, the restriction of $\mathscr{B}$ to $\tilde{\mathfrak{f}}$ is an invariant nondegenerate symmetric billinear form (Lemma 16.22) and $\tilde{\mathfrak{f}}$ is of strictly lower dimension than $\g$.

So now we are back at our beginning situation but with $\g$ replaced by the strictly lower dimension $\tilde{\mathfrak{f}}$. We can therefore iterate this process and after a finite number of steps, splitting off a simple direct summand at each step. Therefore we can decompose $\g$ into the direct sum of simple Lie algebras, so $\g$ is semisimple.$\quad\square$

Dieudonné’s Theorem 16.24 together with Claim 1. of Theorem 16.21 thus show Cartan’s Criterion for Semisimplehood:

Theorem 16.25 (Cartan’s Criterion for Semisimplehood):

A Lie algebra $\g$ is semisimple if and only if all its symmetric invariant billinear forms are also nondegenerate. By Theorem 16.21 symmetric invariant billinear forms exist and are simply related to the Killing form as stated there.$\quad\square$

## The Killing Form as a Compactness Detector Part 2

Now we explore the converse of Theorem 16.12: what does a group’s compactness tell us about the Killing form? Clearly there is no outright converse of Theorem 16.12 for a torus group is a compact Lie group with a Killing form that is identically nought, so compact groups don’t generally have definite Killing forms. From Lemma 16.8, compact groups with continuous centers have degenerate Killing forms, so we shall need to rule these out of any converse theorem statement. The ideas explored in the foregoing section show us that we can study invariant symmetric forms generally and, if we can show nondegeneracy, then all such forms are essentially like the Killing form. Crucial to this study is the idea of invariant averaging in a compact group where we begin with any arbitrary, convenient inner product and average it over the group in a way that is invariant when we displace all the group elements by any constant left or right translation. In doing this, we get a billinear form that is invariant. However, we must (or rather, Alfréd Haar, 1885-1933, did) first come up with the right notion averaging, and this is through the Haar Measure. The group’s compactness is essential here as it lets one to do this averaging with a convergent integral.

We first swiftly review the essential measure and integration concepts needed to explain Haar’s theorem, which tells us that the measure we are about to use is essentially unique. Haar’s theorem applies to locally compact topological groups, of which Lie groups are of course a kind. So we specialize our discussion to Lie groups.

Definition 16.26 ($\sigma$-Algebra):

Let $\mathscr{T}$ be the topology, i.e. the set of all open sets within a Lie group’s group topology. Then we define the $\sigma$-Algebra or Borel Algebra $\mathscr{S}$ of all Borel sets as the set of all sets which can be gotten from $\mathscr{T}$ by a finite sequence of the following operations: Countable union, countable intersection and relative complement (set minus).

Definition 16.27 (Measure):

Let $\mathscr{S}$ be a Lie group’s Borel algebra. Then a measure $\mu:\mathscr{S}\to\R\bigcup\{\infty\}$ is a function from $\mathscr{S}$ to the extended real line fulfilling:

1. $\mu(\mathcal{U})\geq 0,\,\forall \mathcal{U}\in\mathscr{S}$;
2. $\mu(\emptyset)=0$;
3. If $\{\mathcal{U}_k\}_{k=1}^\infty$ is a countable gathering of pairwise disjoint ($k\neq j\Rightarrow\mathcal{U}_j\bigcup\mathcal{U}_k=\emptyset$) Borel sets $\mathcal{U}_k\in\mathscr{S}$ then\begin{equation}\label{CountableAddicitivity}\mu\left(\bigcup\limits_{k=1}^\infty\,\mathcal{U}_k\right) = \sum\limits_{k=1}^\infty \mu\left(\mathcal{U}_k\right)\end{equation}

and now we define the crucial property for our present discussion:

Definition 16.28 (Translationally Invariant Measure):

Let $\mathscr{S}$ be the Borel algebra on a Lie group $\G$. Then a left translationally invariant measure $\mu:\mathscr{S}\to\R\bigcup\{\infty\}$ is a measure on $\mathscr{S}$ such that $\mu(\gamma\,\mathcal{U}) = \mu(\mathcal{U}),\,\forall\,\gamma\in\G,\,\forall\,\mathcal{U}\in\mathscr{S}$. Naturally, a right translationally invariant measure $\mu^\prime$ is one for which $\mu^\prime(\mathcal{U}\,\gamma) = \mu(\mathcal{U}),\,\forall\,\gamma\in\G,\,\forall\,\mathcal{U}\in\mathscr{S}$

and now we are ready to state Haar’s theorem:

Theorem 16.29 ((Haar, 1933)\cite{Haar} Existence and Uniqueness of Translationally Invariant Measure):

Let $\G$ be a Lie group and $\mathscr{S}$ its Borel algebra. Then there exists a unique, up to any real scaling constant, measure $\mu_H:\mathscr{S}\to\R\bigcup\{\infty\}$ on $\mathscr{S}$ fulfilling:

1. $\mu_H$ is left translationally invariant: $\mu_H(\gamma\,\mathcal{U}) = \mu_H(\mathcal{U}),\,\forall\,\gamma\in\G,\,\forall\,\mathcal{U}\in\mathscr{S}$;
2. $\mu_H(\V)$ is finite for any compact set $\V\in\mathscr{S}$;
3. $\mu_H$ is inner regular on $\mathscr{S}$ i.e. $\mu(\U) = \sup \{\mu(\V)|\,\V\subseteq\U,\,\V\in\mathscr{S},\,\V\text{ is compact}\}$;
4. $\mu_H$ is outer regular on $\mathscr{S}$ i.e. $\mu(\U) = \inf \{\mu(\V)|\,\U\subseteq\V,\,\V\in\mathscr{S},\,\V\text{ is open}\}$.

Naturally, the same theorem holds true if “left translationally invariant” is replaced by the analogous notion of right translationally invariant in the above.

Proof: See \cite{Haar}.$\quad\square$

The above theorem holds for any locally compact, second countable topological group. Later versions, particularly by Weil\cite{Weil}, Cartan\cite{Cartan1940} and Alfsen\cite{Alfsen}, of Haar’s theorem removed the second countable assumption, but this is not needed for the Lie group case in the light of Theorem 10.7. If a topological group is locally Euclidean then it is also second countable, and this fact allowed von Neumann\cite{vonNeumann1933} to use the above, second countable, original version of Haar’s theorem to make the first breakthrough on Hilbert’s Fifth Problem and prove it for compact groups. Indeed, von Neumann’s paper is the one straight after Haar’s in the Annals of Mathematics. Now we are ready to prove our partial converse of Theorem 16.12:

Theorem 16.30 (Discrete Centered Compact Groups have Negative Definite Killing Forms):

Let $\G$ be a compact Lie group. Then the Killing form on the Lie algebra $\g$ or $\G$ is negative semidefinite. If, further, the group is centerless or has at most a discrete center, then the Killing form is negative definite.

Proof: Show Proof

Choose any basis for $\g$ and then define an inner product $\langle\cdot,\,\cdot\rangle:\g\times\g\to \R$ to be the one that makes the chosen basis orthonormal. Now let $\mu$ be a right translationally invariant Haar measure on $\G$ with its Borel algebra $\mathscr{S}$; since the group is compact we can define the following billinear form on $\g$ through the convergent integral, defined as the Lebesgue integral taken with respect to the Haar measure.

By construction, $\mathscr{B}$ is $\Ad$-invariant: to calculate $\mathscr{B}(\Ad(\zeta)\,X,\,\Ad(\zeta) Y)$ , we simply make the substitution $\omega =\gamma\, \zeta$ and then the integral becomes $\int_{\omega\in\G}\,\langle\Ad(\omega)\,X,\,\Ad(\omega)\,Y\rangle\, \d\,\mu(\omega\,\zeta^{-1})$ (since $\Ad(\gamma)\,\Ad(\zeta)=\Ad(\gamma\,\zeta)$) and then $\mu(\U\,\zeta^{-1})=\mu(\U)\,\forall\,\U\in\mathscr{S}$ and we see that $\mathscr{B}(\Ad(\zeta)\,X,\,\Ad(\zeta) Y) = \mathscr{B}(X,\,Y)$.$\mathscr{B}$ is also positive definite, as $\mathscr{B}(X,\,X)>0$ if $X\neq\Or$. So its matrix $\mathbf{B}$ is invertible and has all positive eigenvalues. We can therefore chose a basis for $\g$ that puts $\mathbf{B}$ into its canonical form: a diagonal matrix with only $\pm1$ elements along its leading diagonal – there are no noughts because $\mathbf{B}$ is invertible – and in this case all the elements are positive. Therefore in this basis $\mathbf{B}=\id$.

As we have seen, $\Ad$-invariance of $\mathscr{B}$ implies (differential) invariance (Lemma 16.7) and contrariwise, so $\mathscr{B}(\ad(X)\,Y,\,Z) +\mathscr{B}(Y,\,\ad(X)\,Z)=\Or,\,\forall\,X,\,Y,\,Z\in\g$. Thus we have $\ad(X)^T\,\mathbf{B} + \mathbf{B}\,\ad(X)=0$ and, in our canonical basis, $\ad(X)^T=-\ad(X)\,\forall\,X\in\g$. So now the Killing form $\mathscr{K}(X,\,X) = \mathrm{tr}(\ad(X)\,\ad(X)) = -\mathrm{tr}(\ad(X)^T\,\ad(X)) \leq 0$ since $\mathrm{tr}(\ad(X)^T\,\ad(X))$ is the Frobenius norm (the same situation that we had in Example 16.4). This proves the Killing form is negative semidefinite.

Because the Killing form is the negative of the Frobenius norm, it can only attain the value of nought for nonzero $X\in\G$ if $\ad(X)=\Or$, i.e. if $\G$ has a continuous center. Therefore, in the case of a group that has at most a discrete center, the Killing form is negative definite and a for a compact group with negative semidefinite Killing form, the Killing form’s kernel is the Lie algebra’s center: the Abelian ideal $\{X\in\g|\;\ad(X)=\Or\}$. $\quad\square$

It is worth comparing the negative definite form $-\mathscr{B}$ and the negative semidefinite Killing form in Theorem 16.30. As we have seen, all invariant, symmetric, billinear forms are related to one another by a nonsingular, block diagonal matrix, where the blocks are proportional to the identity matrix and compatible with the simple subspaces. If we sift carefully through the proof of Theorem 16.30 and ask how it would change in the non semisimple case, we would find that some of the blocks could be zero blocks, thus allowing for the negative definite $-\mathscr{B}$ and a Killing form that can be zero for nonzero vectors. We have the following logical equivalence by dint of Theorem 16.12 and Theorem 16.30:

Theorem 16.31 (Discrete Centered Groups Compact Iff Killing Form Negative Definite):

A Lie group $\G$ which is either centerless or at most has a discrete center is compact if and only if its Killing form $\mathscr{K}:\g\times\g\to\R$ is negative definite.$\quad\square$

Now at last we can prove the one as yet unproven assertion made after in Lemma 16.2:

Corollary 16.32 (No Positive Definite Killing Form Groups):

There are no Lie groups with positive definite Killing forms.

Proof: Suppose group $\G$ has a positive definite Killing form on its Lie algebra $\g$. Then by Theorem 16.12, the group is compact and has at most a discrete center. So now Theorem 16.30 applies and shows that the Killing form must be negative definite, a contradiction.$\quad\square$

The difference between the negative definite and negative semi-definite cases in Theorem 16.30 motivates the following definition:

Definition 16.33 (Reductive Lie Algebra):

A Lie algebra $\g$ whose image under $\ad$ is semisimple is called reductive. Equivalently, a Lie algebra that is the direct sum $\g=\mathfrak{s}\oplus\mathfrak{a}$ of a semisimple algebra $\mathfrak{s}$ and an Abelian Lie algebra $\mathfrak{a}$ is reductive. Equivalently again: a Lie algebra with degenerate Killing form where the kernel of the Killing form equals the center of the algebra is reductive.

So another way to state Theorem 16.30 is that a compact group has a negative semidefinite Killing form and a reductive Lie algebra.

Unfortunately, the radical, or kernel of the Killing form is in general bigger than the center of the Lie algebra in question. We have the following simple fact:

Lemma 16.34 (Kernel of Killing form an Ideal):

The kernel $\ker(\mathscr{K})$ of the Killing form $\mathscr{K}:\g\times\g\to\mathbb{K}$ is an ideal.

Proof: Let $X,\,Z\in\g$ and $Y\in\ker(\mathscr{K})$, then $\mathscr{K}([Y,\,Z],\,X)=\mathscr{K}(Y,\,[Z,\,X])=0\,\forall\,X\in\g$ so that $[Y,\,Z]\in\ker(\mathscr{K})$$\quad\square$

but the kernel of the Killing form may be rather bigger than the group’s center. Moreover, even though any Abelian ideal must be contained in the Killing form’s kernel (Lemma 16.23), the kernel itself may not be an Abelian ideal. Degeneracy of the Killing form in general is a messy and highly complicated situation and arises in the cases of solvable and nilpotent Lie algebras (the latter a special case of the former). The Levi decomposition shows that any finite dimensional Lie algebra can be decomposed as a semidirect product of a solvable ideal and a semisimple Lie algebra.