5 Representations of Lie groups and Lie algebras - generalities 5.1 Basics 5.3 The adjoint representation

5.2 Standard constructions for representations

We give a list of various constructions with representations of Lie groups, and the analagous constructions for their derivatives.

The standard representation of a linear Lie group $G\subseteq\operatorname{GL}_{n}(\mathbb{C})$ comes from its action on $\mathbb{C}^{n}$ :

	$\displaystyle\rho(g)$	$\displaystyle=g$
	$\displaystyle D\rho(X)$	$\displaystyle=X.$

The direct sum of representations $(\rho_{1},V_{1})$ , $(\rho_{2},V_{2})$ is $(\rho_{1}\oplus\rho_{2},V_{1}\oplus V_{2})$ with derivative

D(\rho_{1}\oplus\rho_{2})=D\rho_{1}\oplus D\rho_{2}.

The determinant representation of $G\subset\operatorname{GL}_{n}(\mathbb{C})$ is $\det:G\to\mathbb{C}^{*}$ which sends $g$ to $\det(g)$ . We have

D\det(X)=\operatorname{tr}(X),

which follows from $\det\exp(tX)=e^{t\operatorname{tr}(X)}$ .

If $(\rho,V)$ is a representation of $G$ , the dual representation $(\rho^{*},V^{*})$ of $(\rho,V)$ is defined by

\left(\rho^{*}(g)(\lambda)\right)(v)=\lambda\left(\rho(g^{-1})(v)\right),

for $\lambda\in V^{*}$ a linear functional on $V$ and $g\in G$ . It has derivative

D\rho^{*}(X)(\lambda)(v)=-\lambda(D\rho(X)v).

Given a basis of $V$ , then the matrix of $\rho^{*}$ with respect to the dual basis is

\rho^{*}(g)=\rho(g)^{-T},

which differentiates to

(D\rho^{*})(X)=-D\rho(X)^{T}.

If $(\rho_{1},V_{1})$ and $(\rho_{2},V_{2})$ are representations of $G$ , then as before the tensor product representation $\rho_{1}\otimes\rho_{2}$ is the representation on $V_{1}\otimes V_{2}$ defined by

(\rho_{1}\otimes\rho_{2})(g)=\rho_{1}(g)\otimes\rho_{2}(g).

Then using the product rule one sees

D(\rho_{1}\otimes\rho_{2})=D\rho_{1}\otimes\mathrm{id}_{V_{2}}+\mathrm{id}_{V_% {1}}\otimes D\rho_{2}.

The symmetric square and alternating square are also as for finite groups. If $(\rho,V)$ is a representation of $G$ then $\operatorname{Sym}^{2}(V)$ has a representation $\operatorname{Sym}^{2}\rho$ :

\operatorname{Sym}^{2}\rho(g)(v_{1}\cdot v_{2})=(\rho(g)(v_{1}))\cdot(\rho(g)(% v_{2})),

and

D\operatorname{Sym}^{2}\rho(X)(v_{1}\cdot v_{2})=D\rho(X)(v_{1})\cdot v_{2}+v_% {1}\cdot D\rho(X)(v_{2}).

Similarly we have a representation $\Lambda^{2}\rho$ on $\Lambda^{2}(V)$ :

\Lambda^{2}\rho(g)(v_{1}\wedge v_{2})=\rho(g)(v_{1})\wedge\rho(g)(v_{2}),

D\Lambda^{2}\rho(X)(v_{1}\wedge v_{2})=D\rho(X)(v_{1})\wedge v_{2}+v_{1}\wedge D% \rho(X)(v_{2}).

We can take tensor/symmetric/alternating products of more than one factor. Suppose $(\rho_{i},V_{i})$ are representations of $G$ .

•

We form the tensor product

$V_{1}\otimes V_{2}\otimes\ldots\otimes V_{l}.$

It is generated by symbols $v_{1}\otimes\ldots\otimes v_{l}$ subject to the multilinear relations, that is, linearity in each slot:

$v_{1}\otimes\ldots\otimes(av_{i}+bv_{i}^{\prime})\otimes\ldots\otimes v_{l}=a(% v_{1}\otimes\ldots\otimes v_{i}\otimes\ldots\otimes v_{l})+b(v_{1}\otimes% \ldots\otimes v_{i}^{\prime}\otimes\ldots\otimes v_{l}).$

One has

$\dim\left(V_{1}\otimes V_{2}\otimes\ldots\otimes V_{l}\right)=\prod_{i=1}^{l}% \dim V_{i}.$

The action of $G$ is as before: for $g\in G$ , $v_{i}\in V_{i}$ ,

$g(v_{1}\otimes\ldots\otimes v_{l})=(gv_{1})\otimes\ldots\otimes(gv_{l}).$

The derivative is, for $X\in\mathfrak{g}$ and $v_{i}\in V_{i}$ ,

$\displaystyle X(v_{1}\otimes\ldots\otimes v_{l})=$ $\displaystyle(Xv_{1})\otimes v_{2}\otimes\ldots\otimes v_{l}$

$\displaystyle+v_{1}\otimes(Xv_{2})\ldots\otimes v_{l}$

$\displaystyle+\ldots$

$\displaystyle+v_{1}\otimes v_{2}\ldots\otimes(Xv_{l}).$

We also write

$V^{\otimes l}=V\otimes\cdots\otimes V.$
•

The $l$ th symmetric power is the space $\operatorname{Sym}^{l}(V)$ generated by symbols $v_{1}\cdots v_{l}$ with linearity in each slot and any permutation of the vectors giving the same element. We have

$\dim\operatorname{Sym}^{l}(V)=\binom{n+l-1}{l}=\binom{n+l-1}{n-1}$

where $\dim V=n$ . Indeed, if $e_{1},\ldots,e_{n}$ is a basis for $V$ then a basis for $\operatorname{Sym}^{l}(V)$ is

$\{e_{i_{1}}\ldots e_{i_{l}}:1\leq i_{1}\leq i_{2}\leq\ldots\leq i_{l}\leq n\}$

from which finding the dimension is a simple counting problem.

As for higher tensor powers, the actions of $G$ and $\mathfrak{g}$ are

$g(v_{1}\cdots v_{l})=(gv_{1})\ldots(gv_{l}),$

and

$X(v_{1}\cdots v_{l})=(Xv_{1})v_{2}\ldots v_{l}+\ldots+v_{1}\ldots v_{l-1}(Xv_{% l}).$
•

The $l$ th alternating power is the space $\Lambda^{l}(V)$ generated by symbols $v_{1}\wedge\cdots\wedge v_{l}$ with linearity in each slot and having the alternating property: for any permutation $\sigma\in S_{l}$ , we have

$v_{\sigma(1)}\wedge\cdots\wedge v_{\sigma(l)}=\epsilon(\sigma)(v_{1}\wedge% \cdots\wedge v_{l}).$

In particular, switching the places of two components reverses the sign, while $v_{1}\wedge\cdots\wedge v_{l}=0$ if two of the vectors coincide (more generally, if they are linearly dependent).

We have

$\dim\Lambda^{l}V=\binom{n}{l}$

where $\dim V=n$ . Indeed, if $e_{1},\ldots,e_{n}$ are a basis for $V$ then a basis for $\Lambda^{l}(V)$ is

$\{e_{i_{1}}\ldots e_{i_{l}}:1<i_{1}<i_{2}<\ldots<i_{l}\leq n\}$

from which finding the dimension is a simple counting problem. In particular, $\Lambda^{n}\mathbb{C}^{n}$ is one-dimensional generated by $e_{1}\wedge\cdots\wedge e_{n}$ .

The representation on $\Lambda^{l}(V)$ is given again as above: for $g\in G$ ,

$g(v_{1}\wedge\ldots\wedge v_{l})=(gv_{1})\wedge\ldots\wedge(gv_{l}),$

while for $X\in\mathfrak{g}$

$X(v_{1}\cdots v_{l})=(Xv_{1})\wedge v_{2}\wedge\ldots\wedge v_{l}+\ldots+v_{1}% \wedge\ldots\wedge v_{l-1}\wedge(Xv_{l}).$

To give an example of how to justify the claims about derivatives, we do the case of tensor products. Suppose $V$ , $W$ are vector spaces acted on by $G$ . Let $X\in\mathfrak{g}$ , $v\in V$ , $w\in W$ . We must compute

\left.\frac{d}{dt}\exp(tX)v\otimes\exp(tX)w\right|_{t=0}.

Expanding:

	$\displaystyle\exp(tX)v\otimes\exp(tX)w$	$\displaystyle=(v+tXv+O(t^{2}))\otimes(w+tXw+O(t^{2}))$
		$\displaystyle=v\otimes w+t(Xv\otimes w+v\otimes Xw)+O(t^{2}).$

This gives

	$\displaystyle X(v\otimes w)$	$\displaystyle=\left.\frac{d}{dt}\exp(tX)v\otimes\exp(tX)w\right\|_{t=0}$
		$\displaystyle=Xv\otimes w+v\otimes Xw$

as required.

Remark 5.8.

We defined tensor products (and so on) of representations of Lie groups and then differentiated them. We could also directly make these definitions with Lie algebras. For instance, if $\mathfrak{g}$ is a Lie algebra and $V$ is a representation of $\mathfrak{g}$ , we define the symmetric square representation on $\operatorname{Sym}^{2}(V)$ by

X(vw)=(Xv)w+v(Xw).

5.2.1 Functional constructions

We can construct representations as vector spaces of functions on topological spaces with actions of $G$ . If $G$ acts on a set $X$ , then it also acts on the vector space of functions $X\rightarrow\mathbb{C}$ by $(g\cdot f)(x)=f(g^{-1}x)$ . Usually this will be infinite dimensional, and so out of the scope of our course, but sometimes we can impose conditions allowing us to handle it. For example, $\operatorname{GL}_{n}(\mathbb{C})$ acts on $\mathbb{C}^{n}$ , and hence on the space of polynomial functions in $n$ variables. Imposing a further restriction — to homogeneous polynomials of some fixed degree — gives a finite-dimensional representation. The derivative must be calculated on a case-by-case basis. We will see examples of this on the problem sheet.

	$\displaystyle X(v_{1}\otimes\ldots\otimes v_{l})=$	$\displaystyle(Xv_{1})\otimes v_{2}\otimes\ldots\otimes v_{l}$
		$\displaystyle+v_{1}\otimes(Xv_{2})\ldots\otimes v_{l}$
		$\displaystyle+\ldots$
		$\displaystyle+v_{1}\otimes v_{2}\ldots\otimes(Xv_{l}).$