Torus Actions

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Given an arbitrary finite group $G$, one way to examine it closely is through its finite-dimensional representation

\[\rho:G\rightarrow \Aut(V)\]

Once we choose a basis for $V$, analyzing the image of $G$ under $\rho$ becomes purely a matter of linear algebra, allowing us to discern the structure of $G$ much more easily.

For Lie groups, this representation-theoretic perspective is even more useful, because Lie groups inherently act on other objects such as $\GL(n;\mathbb{R})$ or $\Diff(M)$.

However, as in §Representation Theory of Finite Groups, ⁋Definition 1, if we define the representation theory of $G$ in this way, we lose the smooth structure on the Lie group $G$. Thus we must modify the definition as follows.

Definition 1 For a Lie group $G$, a representation of $G$ consists of a finite-dimensional vector space $V$ together with a smooth map

\[\rho:G\rightarrow \Aut(V)\]

If we regard $G$ as a Lie group equipped with the discrete topology and the trivial smooth structure, then this definition can be viewed as a generalization of §Representation Theory of Finite Groups, ⁋Definition 1. Similarly, all definitions in §Representation Theory of Finite Groups, §§Basic Concepts of Representation Theory can be carried over to Lie groups.

What played a crucial role in the preceding discussion was the fact that $G$ is a finite group. For instance, the idea of averaging over all elements of $G$ relied on this. To generalize this to Lie groups, we must impose some kind of finiteness condition on $G$.

Therefore, we often consider the case where $G$ is a compact Lie group. In this case, as a locally compact Hausdorff space, $G$ admits a Haar measure, and thus we can replace sums over elements of $G$ with integrals over all of $G$. Of course, this requires properly defining $\delta_x$ functions and generalizing function spaces appropriately, but we omit these details as they are not our present objective. What matters is that the representation theory of Lie groups can be approached by the same methodology as that of finite groups. In particular, any finite-dimensional representation decomposes as a direct sum of irreducible representations.

On the other hand, given a finite-dimensional representation $G\rightarrow\Aut(V)$, the greatest advantage is that we can regard the images $\rho(g)$ as matrices (via a choice of basis). Hence we can investigate them using our tools for matrices and linear maps.

One of the most important tools in linear algebra is diagonalization. Thus, for a given Lie group action $\rho:G \rightarrow \Aut(V)$, we are interested in choosing a basis of $V$ so that the matrix representation of each $\rho(g)$ becomes diagonal. If $G$ were finite, we could find such a basis for each $g$ individually, but since $G$ is now infinite, this is difficult. Therefore, we naturally become interested in elements that are simultaneously diagonalizable. Now, §Eigenspace Decomposition, ⁋Proposition 10 tells us that two diagonalizable matrices are simultaneously diagonalizable if and only if they commute, so the following definition is appropriate.

Definition 2 For a compact, connected Lie group $G$, a subgroup $T$ of $G$ is called a maximal torus if $T$ is a torus and is maximal with respect to inclusion.

Choosing an arbitrary element $X$ of the Lie algebra $\mathfrak{g}$, the closure of the one-parameter subgroup traced by the exponential map in this direction becomes a torus. Hence the existence of a maximal torus is immediate by §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma).

It is worth noting that one might be tempted to think that the one-parameter group $\exp(tX)$ above always produces a one-dimensional torus, but this is not always the case. For example, consider the two-dimensional torus

\[T^2\cong \mathbb{R}^2/\mathbb{Z}^2\]

Before taking the quotient, the one-parameter subgroup in the direction $(1,\sqrt{2})$ densely covers all of $T^2$, and the closure of its image is exactly all of $T^2$. An element $t\in T$ satisfying

\[T=\overline{\langle t\rangle}\]

is called a generator of $T$.

Weight decomposition

Our claim is that $\rho(T)$ is simultaneously diagonalizable. To show this, it suffices to prove that each element of $\rho(T)$ is diagonalizable. To this end, consider an arbitrary finite-dimensional representation $\rho:G\rightarrow \Aut(V)$ and its restriction $\rho\vert_T$ to a maximal torus $T$. The first thing to check is that $T$ is a compact Lie group. Therefore, both §Representation Theory of Finite Groups, ⁋Proposition 6 and §Representation Theory of Finite Groups, ⁋Lemma 8 (Schur) hold.

Writing these out in more detail, first from the fact that $\rho\vert_T$ is a unitary representation, we obtain the following irreducible decomposition

\[V=\bigoplus_i V_i\]

Here each $V_i$ is an irreducible $T$-representation. On the other hand, since $T$ is abelian, for any $t\in T$, $\rho(t)$ commutes with the $T$-action, and thus the restriction of $\rho(t)$ to each $V_i$ is a $T$-automorphism. Now by the second result of §Representation Theory of Finite Groups, ⁋Lemma 8 (Schur), $\rho(t)$ acts by a scalar multiple

\[\rho(t)(v)=\lambda_i(t)v\qquad \lambda_i(t)\in \mathbb{C}^\times\]

But now if $T$ acts on $V_i$ by scalar multiples, then any subspace of $V_i$ would be $T$-invariant, and therefore for $V_i$ to be irreducible we must have $\dim V_i=1$.

Now since $\dim V_i=1$, we have $\Aut(V_i)\cong \mathbb{C}^\times$, and we see that the above $\lambda_i: T\rightarrow \mathbb{C}^\times$ corresponds exactly to a character of $\rho$, so the irreducible decomposition can be thought of as being directly parametrized by characters $\lambda_i$. That is, consider the irreducible decomposition given by the formula

\[V=\bigoplus_\lambda V_\lambda;\qquad V_\lambda=\{t\cdot v=\lambda(t)v\text{ for all $t\in T$}\}\]

Then for each $t\in T$, $\rho(t)$ is diagonalized by this decomposition, and the eigenvalue corresponding to each eigenspace $V_\lambda$ is $\lambda(t)$. A different choice of $t$ leaves the above decomposition unchanged and only changes the eigenvalue assigned to each eigenspace $V_\lambda$.

Intuitively, we can think of $t\mapsto e^{2\pi i\lambda_i(X)}$ as a rotational motion with angular velocity $\lambda_i(X)$, and adopting this perspective, we know that for each $X\in \mathfrak{t}$, we can describe this torus action by the angular velocity $\lambda_i(X)$ in this direction. Each of these $\lambda_i$ is called a weight. Then for each weight $\lambda_i$, there exists an appropriate $V_i$ on which the torus action operates as $t\cdot v=\rchi_{\lambda_i}(t)v$. Such a $V_i$ is called a weight space.

Example 3 As a concrete example, consider the one-dimensional torus

\[S^1\cong T \cong \mathbb{R}/\mathbb{Z}\]

Then $S^1$ can be identified with the set

\[S^1=\left\{e^{2\pi i t}\mid t\in \mathbb{R}/\mathbb{Z}\right\}\]

Now suppose this set acts on the two-dimensional vector space $\mathbb{C}^2$ by the formula

\[e^{2\pi i t}\cdot (z_1,z_2)=(e^{4\pi i t}z_1, e^{-2\pi i t}z_2)\]

This action may look artificial, but as we saw above, given any torus $T$ and any representation $V$, if we consider the irreducible decomposition of $V$ and choose bases $e_i$ for each irreducible component, then every torus action can be written in this form (by an appropriate choice of basis).

Writing this as a matrix, the above action is represented by the element (family of elements) of $\GL(2;\mathbb{C})$

\[\begin{pmatrix}e^{4\pi i t}&0\\0&e^{-2\pi i t}\end{pmatrix}\]

At this point, the trace $e^{4\pi i t}+e^{-2\pi i t}$ of this matrix is precisely the character of this representation.

The weight spaces of this action are obviously $\span(e_1), \span(e_2)$, and for example the weight corresponding to $\span(e_1)$ is given by the linear functional $\lambda_1:\mathfrak{t}\rightarrow \mathbb{C}$ satisfying

\[\rchi_{\lambda_1}(\exp (X))=e^{2\pi i \lambda_1(X)}\qquad\text{for all $X\in \mathfrak{t}$}\]

This is of course defined by $\lambda_1(t)=2t$, which sends $1\in \mathbb{R}$ to $2$, and thus the weight corresponding to this weight space can be said to be $2$ (with a slight abuse of notation). For $\lambda$ to satisfy the above formula, since $e^{2\pi i}=1$, we must have $\lambda(1)\in \mathbb{\mathbb{Z}}$.

More generally, if an action of an $r$-dimensional torus is given, then $\mathfrak{t}$ will be $\mathbb{R}^r$, and if we write the torus $T$ as

\[T^r=\left\{(e^{2\pi i t_1}, \ldots e^{2\pi i t_r})\mid t_i\in \mathbb{R}/\mathbb{Z}\right\}\]

then the elements that can be weights in its Lie algebra $\mathfrak{t}\cong \mathbb{R}^r$ are those belonging to $\mathbb{Z}^r$, and thus a weight $\lambda$ will be given by the following $r$-tuple

\[\lambda=(n_1, \ldots, n_r)\]

Explicitly, this weight is the linear functional that outputs $n_1x_1+\cdots+n_rx_r$ when given arbitrary $X=(x_1,\ldots, x_r)\in \mathfrak{t}$.

Just as in eigenspace decomposition in linear algebra, the multiplicity of each weight need not be $1$. For example, consider the following torus action

\[e^{2\pi i t}\cdot(z_1, z_2)=(e^{4\pi i t}z_1, e^{4\pi i t} z_2)\]

then this time $T$ acts on the two-dimensional space $\mathbb{C}^2$ as if it had weight $2$. Collecting components that have the same weight $\lambda$ in this way and calling this $V_\lambda$, we obtain the weight space decomposition $V=\bigoplus V_\lambda$. Writing the discussion so far rigorously as a definition, we have the following.

Definition 4 Let a torus $T$ and a complex $T$-module $V$ be given. For an irreducible character $\rchi_\lambda: T \rightarrow S^1$ and the corresponding linear functional $\lambda:\mathfrak{t}\rightarrow\mathbb{C}$, we say that $\lambda$ is a weight of $V$ if the set

\[V_\lambda=\left\{v\in V\mid t\cdot v=\rchi_\lambda(t)v\text{ for all $t\in T$}\right\}\]

is nontrivial. In this case, we call $V_\lambda$ the weight space of $\lambda$, and the decomposition

\[V=\bigoplus_\lambda V_\lambda\]

is called the weight decomposition of $V$.

Maximal tori

In this section we show that any element of a compact connected Lie group $G$ is always contained in some maximal torus, and that all maximal tori are conjugate to each other.

Our claim is that for a compact connected Lie group $G$ and its maximal torus $T$, the map

\[q: G/T\times T\rightarrow G; \qquad (g,t)\mapsto gtg^{-1}\]

is surjective. Then in particular, for any other torus $T'$ and its generator $t'$, we would be able to find $t\in T$ satisfying $gtg^{-1}=t'$, and using the maximality of $T$ and $T'$ respectively, the two claims mentioned above would be proved.

We can not only show that this map is surjective, but also compute its explicit mapping degree. For this purpose, let us define the following.

Definition 5 For a compact, connected Lie group $G$, a maximal torus $T$, and the normalizer of $T$

\[N=\{g\in G\mid gTg^{-1}=T\}\]

we define the group $W=N/T$ as the Weyl group of $G$.

Then the following holds.

Proposition 6 The Weyl group $W=N/T$ is always finite.

Proof

By definition, $N$ acts on $T$ by conjugation

\[N\rightarrow\Aut(T);\qquad n\mapsto (t\mapsto ntn^{-1})\]

However, $\Aut(T)$ is determined by where the lattice of the torus $T=\mathbb{R}^k/\mathbb{Z}^k$ is sent, and this is embedded in $\GL(k;\mathbb{Z})$ via $\Ad(n)$. That is, this action can be thought of as a continuous map from $N$ to $\GL(k;\mathbb{Z})$. But since $\GL(k;\mathbb{Z})$ is discrete, considering the identity component $N_0$ of $N$, $N_0$ must all be sent to the identity matrix. That is, $N_0$ acts trivially on $T$.

Now for an arbitrary one-parameter subgroup $\alpha:\mathbb{R}\rightarrow N_0$, we must have $\alpha(\mathbb{R})\cdot T=T$, and from this we know that $\alpha(\mathbb{R})\subset T$. But by §Vector Fields, ⁋Theorem 6, these cover some open neighborhood of the identity in $N_0$, and therefore generate $N_0$. That is, $N_0=T$.

Therefore $N/T$ is exactly the number of connected components of $N$, and since $N$ is a closed subspace of the compact Lie group $G$, it is likewise compact, so this cannot be infinite.

Now our claim is as follows.

Lemma 7 For a compact, connected Lie group $G$, a maximal torus $T$, and the map

\[q:G/T\times T\rightarrow G;\qquad (gT, s)\mapsto gsg^{-1}\]

the mapping degree of $q$ is $\lvert W\rvert$.

Proof

To compute the mapping degree, it suffices to choose one regular value and find all its preimages, then compute the sign of the differential at each preimage.

For this, first choose a generator $t$ of $T$ and consider its preimage $q^{-1}(t)$. For arbitrary $(gT,s)\in G/T\times T$, the condition $q(gT,s)=t$ means $gsg^{-1}=t$. On the other hand, $g^{-1}tg=s\in T$, and

\[\overline{\langle s\rangle}=g^{-1}\overline{\langle t\rangle}g=g^{-1}Tg\]

but since $s\in T$, we have $g^{-1}Tg\subseteq T$. However, since conjugation is a homeomorphism, $g^{-1}Tg$ is a torus isomorphic to $T$, and from this we know that $g^{-1}Tg=T$ and $g\in N=N_G(T)$. Also, since $s=g^{-1}tg=(gT)\cdot t$,

\[q^{-1}(t)=\{(gT, (gT)\cdot t)\mid gT\in W\}\]

Here $W$ is the Weyl group defined as $N/T$, and we see that $q^{-1}(t)$ is in one-to-one correspondence with $W$.

Therefore, what remains is to show that all of these have the same sign, so that the mapping degree comes out to be exactly $\lvert W\rvert$. For this, we identify the tangent spaces of $G/T\times T$ and $G$ with $\mathfrak{g}$. Specifically, decomposing the Lie algebra $\mathfrak{g}$ into the direct sum of $\mathfrak{t}$ and its orthogonal complement $\mathfrak{f}=\mathfrak{t}^\perp$

\[\mathfrak{g}=\mathfrak{t}\oplus\mathfrak{f}\]

then near the origin, the tangent space of $T$ is $\mathfrak{t}$ and the tangent space of $G/T$ is $\mathfrak{f}$. Thus the tangent space of $G/T\times T$ is $\mathfrak{t}\oplus\mathfrak{f}\cong\mathfrak{g}$.

On the other hand, for arbitrary $X\in\mathfrak{t}$ and $Y\in\mathfrak{f}$, the differential of $q$ at $(eT,t)$ is computed as follows. For the direction $X$, that is, considering variation in the $T$ direction,

\[d q_{(eT,t)}(X,0)=\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}q(eT, t\exp(\epsilon X))=\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}t\exp(\epsilon X)=X\]

Here we used that $T$ is abelian, so $t$ and $\exp(\epsilon X)$ commute. Next, for the direction $Y$, that is, considering variation in the $G/T$ direction,

\[d q_{(eT,t)}(0,Y)=\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}q(\exp(\epsilon Y)T, t)=\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}\exp(\epsilon Y)t\exp(-\epsilon Y)\]

Now writing $t=\exp(H)$ ($H\in\mathfrak{t}$),

\[\exp(\epsilon Y)t\exp(-\epsilon Y)=\exp(\epsilon Y)\exp(H)\exp(-\epsilon Y)=\exp(\Ad_{\exp(\epsilon Y)}(H))=\exp(e^{\epsilon\ad_Y}H)\]

and therefore

\[\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}\exp(\epsilon Y)t\exp(-\epsilon Y)=\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}\exp(H+\epsilon[Y,H])=\exp(H)\cdot [Y,H]\]

Here since $\mathfrak{t}$ is abelian, $[Y,H]\in\mathfrak{f}$, and since $\exp(H)=t$, this can be written as $t\cdot(\Ad_t^{-1}(Y)-Y)$. Summarizing, under appropriate identification,

\[d q_{(eT,t)}=\begin{pmatrix} I & 0 \\ 0 & \Ad_t^{-1}\vert_\mathfrak{f}-I \end{pmatrix}\]

Here the first block corresponds to the $\mathfrak{t}$ direction and the second block to the $\mathfrak{f}$ direction.

Now we show that $\Ad_t^{-1}\vert_\mathfrak{f}-I$ is invertible, and that the signs match at all preimages. If there exists $Y\in\mathfrak{f}$ such that $(\Ad_t^{-1}-I)Y=0$, then $\Ad_t(Y)=Y$. Then for any integer $m$, we have $\Ad_{t^m}(Y)=Y$, and from the assumption that $t$ is a generator, we have $\Ad_s(Y)=Y$ for all $s\in T$. Now for arbitrary $H\in\mathfrak{t}$,

\[[H,Y]=\frac{d}{d\epsilon}\bigg\vert_{\epsilon=0}\Ad_{\exp(\epsilon H)}(Y)=0\]

so $Y$ commutes with all elements of $\mathfrak{t}$. But since $\mathfrak{t}$ is a maximal abelian subalgebra, $Y\in\mathfrak{t}$, and therefore $Y\in\mathfrak{f}\cap\mathfrak{t}=\{0\}$. That is, $\Ad_t^{-1}\vert_\mathfrak{f}-I$ is invertible.

Finally, let us verify that the determinant of $dq$ has the same sign at all points of $q^{-1}(t)$. Choose arbitrary $w\in W$ and let it be represented by $x\in N$. Then since $q(xT,x^{-1}tx)=t$, we have $(xT, x^{-1}tx)\in q^{-1}(t)$. To compute the differential at this point, from the definition of $q$,

\[q(gT, s)=gsg^{-1}\]

so, considering left translation by $x$ and conjugation by $x$,

\[d q_{(xT, x^{-1}tx)}=\Ad_x\circ d q_{(eT, t)}\circ (\text{left translation})\]

In particular, both $\Ad_x\vert_\mathfrak{f}$ and $\Ad_x\vert_\mathfrak{t}$ are linear maps with determinant $1$ (the former because it is an orthogonal map, the latter because $x\in N$ so $\Ad_x$ preserves $\mathfrak{t}$), and therefore the determinant of $d q_{(xT, x^{-1}tx)}$ equals the determinant of $d q_{(eT,t)}$.

On the other hand, $\det(\Ad_t^{-1}\vert_\mathfrak{f}-I)$ is also the same for $w\cdot t$. Indeed,

\[\Ad_{wt^{-1}w^{-1}}\vert_\mathfrak{f}-I=\Ad_w\circ(\Ad_t^{-1}\vert_\mathfrak{f}-I)\circ\Ad_w^{-1}\]

so these two operators are similar and thus have the same determinant.

From the above, we have verified that $t$ is a regular value of $q$, the number of elements in $q^{-1}(t)$ is $\lvert W\rvert$, and the determinant of $dq$ has the same sign at all preimage points. Therefore, with an appropriate choice of orientation, $\deg q=\lvert W\rvert$.

As mentioned above, the core content of this section follows immediately from this lemma.

Theorem 8 For a compact connected Lie group $G$, the following hold.

Any element of $G$ is contained in some maximal torus.
Any two maximal tori of $G$ are conjugate.

Therefore, for any compact connected Lie group $G$ and maximal torus $T$, we obtain the decomposition

\[G=\bigcup_{g\in G}gTg^{-1}\]

This is called the Cartan decomposition.

Weyl group parametrization

The Cartan decomposition tells us that each element of $G$ belongs to some maximal torus, but we can describe this decomposition more explicitly. The key is that the map

\[q:G/T\times T\rightarrow G;\qquad (gT,t)\mapsto gtg^{-1}\]

defined in Lemma 7 is a $\lvert W\rvert$-to-1 covering. From this, each element of $G$ has $\lvert W\rvert$ preimages, and the relations among them are exactly described by the Weyl group.

Specifically, define the following $W$-action on $G/T\times T$:

\[w\cdot(gT, t)=(gw^{-1}T, wtw^{-1})\]

Then

\[q(w\cdot(gT,t))=q(gw^{-1}T, wtw^{-1})=gw^{-1}(wtw^{-1})wg^{-1}=gtg^{-1}=q(gT,t)\]

so $q$ is $W$-invariant, and thus induces a function from the orbit space $(G/T\times T)/W$ to $G$

\[(G/T\times T)/W\rightarrow G\]

Lemma 7 proves that this function is a bijection.

On the other hand, let $\Conj(G)$ denote the space of conjugacy classes of $G$. Then each conjugacy class

\[[g]=\{hgh^{-1}\mid h\in G\}\]

is described through $T$ and $W$ as follows.

Proposition 9 Two elements of a maximal torus $T$ are conjugate in $G$ if and only if they belong to the same orbit under the Weyl group action.

Proof

Let $x,y$ be two elements of $T$ that are conjugate to each other. That is, for some $g\in G$, we have $gxg^{-1}=y$. Now comparing $T$ and $gTg^{-1}$, these are maximal tori in the centralizer $Z_G(y)$ of $y$. Therefore, there exists $h\in Z_G(y)$ such that $T=h(gTg^{-1})h^{-1}$, and from this $(hg)x(hg)^{-1}=y$ and $hg\in N_G(T)$. That is, $y=(hgT)\cdot x$, so $x$ and $y$ belong to the same $W$-orbit.

Conversely, if $x$ and $y$ belong to the same $W$-orbit, they are obviously conjugate in $G$.

From this, we know that for each conjugacy class $[g]$, the intersection $[g]\cap T$ is exactly one $W$-orbit. Therefore we obtain the following one-to-one correspondence.

Proposition 10 There is a natural one-to-one correspondence between $T/W$ and $\Conj(G)$.

Under this one-to-one correspondence, the conjugation action on $G$ acts trivially on $T/W$. That is, each element of $G$ preserves the conjugacy class to which it belongs.

Decomposition of the conjugation action

Now let us examine the situation where $G$ acts on itself by conjugation. For this, for each $h\in G$, define the conjugation map

\[c_{h}:G\rightarrow G;\qquad g\mapsto hgh^{-1}\]

Our goal is to transfer this to $(G/T\times T)/W$ and see concretely how it acts.

First, observe the following. For arbitrary $(gT, t)\in G/T\times T$ and $h\in G$,

\[c_{h}(gtg^{-1})=h(gtg^{-1})h^{-1}=(hg)t(hg)^{-1}=q(hgT, t)\]

holds. That is, when we pull back $c_h$ to $G/T\times T$ via the identification $(G/T\times T)/W\cong G$, $c_h$ sends $(gT, t)$ to $(hgT, t)$, so we can think of the following $G$-action being defined on $G/T\times T$:

\[h\cdot(gT,t)=(hgT,t)\]

Now it is obvious that this action commutes with the $W$-action, and therefore $G$ induces a well-defined action on the quotient $(G/T\times T)/W$ as well. From this observation we obtain the following.

Proposition 11 Under the identification $(G/T\times T)/W\cong G$, the conjugation action is expressed as follows.

\[h\cdot[(gT,t)]=[(hgT,t)]\]

That is, it acts by left multiplication on the $G/T$ component and preserves the $T$ component.

On the other hand, by Proposition 10, we know that there is a one-to-one correspondence between $T/W$ and $\Conj(G)$. By definition, $c_h$ does not change the conjugacy class of $G$, and we can check that this is reflected in the fact that there is no change in the $T$ direction in the above proposition. Instead, the conjugation action can be thought of as acting exactly on $G/T$.

To understand the $G$-action on $G/T$ in more detail, we will rewrite this space using the Weyl group. First, for $N=N(T)$, consider the following projection map

\[\pi: G/T\rightarrow G/N;\qquad gT\mapsto gN\]

Then for each coset $gN\in G/N$, its fiber is

\[\pi^{-1}(gN)=\{hT\mid h\in gN\}=\{gxT\mid x\in N\}\]

and since $g$ is fixed, this fiber is essentially $\{xT\mid x\in N\}$, that is, the same as $N/T$. Furthermore, topologically we can verify that $\pi$ is exactly a $\lvert W\rvert$-fold covering map. (§Covering Spaces, ⁋Definition 3) More precisely, this is a principal $W$-bundle with each fiber being $W$.

Example: $\SU(2)$

Let us verify the discussion so far in the compact connected Lie group

\[\SU(2)=\{A\in\GL(2;\mathbb{C})\mid A^\dagger A=I,\det A=1\}=\left\{\begin{pmatrix}\alpha&-\overline{\beta}\\\beta&\overline{\alpha}\end{pmatrix}\,\middle\vert\;\alpha,\beta\in \mathbb{C},\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1\right\}\]

First, we need to find a maximal torus of $\SU(2)$. Our claim is that the set

\[T=\left\{\begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}\,\middle\vert \;\theta\in\mathbb{R}/2\pi\mathbb{Z}\right\}\]

is a (particular) maximal torus of $\SU(2)$. That $T$ is a one-dimensional torus is obvious, so it suffices to show maximality. For this, suppose $A$ is an abelian subgroup of $\SU(2)$ containing $T$. Then any element

\[\begin{pmatrix}a&b\\c&d\end{pmatrix}\in A\]

must commute with all elements of $T$, and in particular must commute with

\[\begin{pmatrix}i&0\\0&-i\end{pmatrix}\]

Computing this,

\[\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}i&0\\0&-i\end{pmatrix}=\begin{pmatrix}ai&-bi\\ci&-di\end{pmatrix},\qquad \begin{pmatrix}i&0\\0&-i\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}ai&bi\\-ci&-di\end{pmatrix}\]

so from this we know that $b=c=0$. From this we know that $A\subset T$.

Now to compute the Weyl group, let us show that the normalizer $N=N_{\SU(2)}(T)$ of $T$ is given by the formula

\[N=T\cup \begin{pmatrix}0&1\\-1&0\end{pmatrix}T\]

First, for arbitrary $g\in \SU(2)$, writing

\[g=\begin{pmatrix}a&b\\c&d\end{pmatrix}\]

we have

\[g \begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}g^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \begin{pmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{pmatrix} = \begin{pmatrix} \lvert\alpha\rvert^2 e^{i\theta} + \lvert b\rvert^2 e^{-i\theta} & a\overline{c}e^{i\theta} + b\overline{d}e^{-i\theta} \\ c\overline{a}e^{i\theta} + d\overline{b}e^{-i\theta} & \lvert c\rvert^2 e^{i\theta} + \lvert d\rvert^2 e^{-i\theta} \end{pmatrix}\]

so for this to belong to $T$, for all $\theta$ we must have

\[a\overline{c}e^{i\theta}+b\overline{d}e^{-i\theta}=0\]

so $a\overline{c}=0$ and $b\overline{d}=0$. If $a\neq 0$, then $\overline{c}=0$ so $c=0$, and from $ad-bc=1$ we get $d=\overline{a}$, and thus from $b\overline{d}=0$ we must have $b=0$. That is, in this case $g\in T$. If $a=0$, then since $\lvert a\rvert^2+\lvert b\rvert^2=1$ we have $\lvert b\rvert=1$, and from $ad-bc=1$ we must have $c=\overline{b}$, and similarly $d=0$. From this we know that $g$ must be of the form

\[\begin{pmatrix}0&e^{i\alpha}\\ -e^{-i\alpha}&0\end{pmatrix}\]

so we can verify that $N$ is given by the above formula. Moreover, the nontrivial element of the Weyl group $W\cong\mathbb{Z}_2$ corresponds to $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, and the action of this element on $T$ is

\[\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}e^{-i\theta}&0\\0&e^{i\theta}\end{pmatrix}\]

so it corresponds to the reflection $\theta\mapsto -\theta$ on the torus $S^1$.

Now considering the standard representation of $\SU(2)$ on $\mathbb{C}^2$, we have $\Aut(\mathbb{C}^2)=\GL(2;\mathbb{C})$, and thus the representation $\rho:\SU(2)\rightarrow \GL(2;\mathbb{C})$ is the inclusion. The same holds for its restriction to the torus $T$, and from this we know that the weights of $T$ are $\theta, -\theta$, the corresponding weight spaces are $\mathbb{C}e_1, \mathbb{C}e_2$, and the weight decomposition is $\mathbb{C}e_1\oplus \mathbb{C}e_2$. On this, we can verify that the Weyl group acts by interchanging the weights.

Weyl chamber

Finally, we examine a definition that gives a taste of the motivation for root systems, which we will cover in the next post. In Lemma 7, we computed the preimage of $q$ at a generator $t$ of $T$ and checked that there are $\lvert W\rvert$ of them. Now that we know $W$ acts on $T$, we can write this condition as follows.

Definition 12 An element $t$ of a maximal torus $T$ is called regular if the only $w\in W$ satisfying $wtw^{-1}=t$ is $w=e$. Conversely, if there exists $w\neq e$ such that $wtw^{-1}=t$, then $t$ is called singular.

That is, a regular element is one whose stabilizer under the Weyl group action is trivial, and a singular element is one with a nontrivial stabilizer. In general, to compute the mapping degree of $q$, one must compute the value at a regular value of $q$, and thinking of the orbit-stabilizer theorem, we know that for the preimage to have $\lvert W\rvert$ images in this way, the stabilizer must necessarily be trivial. From this, it is natural to call such elements regular.

Example 13 In the case of $\SU(2)$, $T=\{\text{diag}(e^{i\theta}, e^{-i\theta})\}$ and $W=\mathbb{Z}_2$ acts by $\theta\mapsto -\theta$. Therefore:

Regular: Elements with $\theta \neq 0, \pi$. These are not fixed points of the reflection.
Singular: $\theta=0$ (identity) and $\theta=\pi$ ($\text{diag}(-1,-1)$). These are fixed by the reflection.

On the torus $T\cong S^1$, there are exactly two singular elements, and the regular elements are their complement.

In general, the following holds.

Proposition 14 For a maximal torus $T$ of a compact connected Lie group $G$:

The regular elements form a dense open subset of $T$.
The singular elements form a closed subset of codimension $\geq 1$ in $T$.
The set of singular elements is a union of finitely many subgroups.

Proof

(1) and (2): For each $w\in W$, $w\neq e$, the fixed point set $\{t\in T: wtw^{-1}=t\}$ is a proper closed subgroup of $T$. The set of singular elements is the finite union of these, so it is closed, and its complement (the regular elements) is dense open.

(3): The fixed point set for each $w\neq e$ is a closed subgroup of $T$, and since $W$ is finite, it is a union of finitely many subgroups.

The set formed by singular elements in $T$ divides the torus into several pieces. For example, in the case of $\SU(2)$, the two singular elements divide $S^1$ into two semicircles. On each semicircle, the Weyl group defines a free action, and the Weyl group action serves to identify them with each other.

From this perspective, the quotient $T_{\text{reg}}/W$ of regular elements of $T$ is a connected space, and this can be connected to the concept of a Weyl chamber. In the case of $\SU(2)$, $T_{\text{reg}}/W \cong (0,\pi)$ is a one-dimensional interval, and this corresponds exactly to a one-dimensional Weyl chamber.

For more general Lie groups, singular elements divide the torus into several chambers, and each chamber serves as a fundamental domain for the Weyl group action. This is the geometric foundation of root systems and Weyl chambers, which we will cover in the next post.

References

[Bro] Armand Borel, Linear Algebraic Groups, Graduate texts in mathematics, Springer, 1991.
[BtD] Theodor Bröcker, Tammo tom Dieck, Representations of Compact Lie Groups, Graduate texts in mathematics, Springer, 1985.

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K

Torus Actions

Weight decomposition

Maximal tori

Weyl group parametrization

Decomposition of the conjugation action

Example: \(\SU(2)\)

Weyl chamber

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