The Lie group ${\rm SU}(N)$ is connected and compact, therefore the exponential map is surjective. In other words, if $g\in {\rm SU}(N)$ there is $X\in {\frak su}(N)$ such that $g = \exp X$.

Physicists often exploit this to turn the problem of finding unitary representations of ${\rm SU}(N)$ in terms of anti-hermitian representations of $\mathfrak{su}(N)$. In that case if ${\bar D}:\mathfrak{su}(N)\to {\operatorname{End}}(V)$ is one anti-hermitian representation of $\mathfrak{su}(N)$ they define $D : {\rm SU}(N)\to \operatorname{GL}(V)$ by $$D(\exp X)=\exp {\bar D}(X)\tag{1}.$$

Now, I have a problem with this. Since the exponential map is continuous, and in this case it is surjective, if it were injective it would give a homeomorphism between ${\frak su}(N)$ and ${\rm SU}(N)$. This cannot happen since ${\frak su}(N)$ is non-compact. Therefore the exponential map cannot be injective.

But this makes (1) ambiguous. The reason is that given $g\in {\rm SU}(N)$ there is not just one $X\in \mathfrak{su}(N)$ with $\exp X =g$, but there may be more. Say there are $X_1,\dots, X_n \in \exp^{-1}(g)$, then it is not clear which one we should pick to use (1), unless of course it were the case that ${\bar D}(X_i) = {\bar D}(X_j)$ for all $X_i,X_j \in \exp^{-1}(g)$ for all $g\in {\rm SU}(N)$, which I can't see why would be true for general ${\bar D}$.

In that case why is it ok to use (1) to define one ${\rm SU}(N)$ representation in terms of one ${\frak su}(N)$ representation? What happens with this injectivity issue I have described?


You are correct that there's an ambiguity in general; e.g. this ambiguity exists for $SO(N)$ and is why the spin representations exist. For $SU(N)$ the ambiguity never occurs because it's simply connected. More generally, we have the following:

Proposition 1: If $G$ and $H$ are connected Lie groups, then the differentiation map $\text{Hom}(G, H) \to \text{Hom}(\mathfrak{g}, \mathfrak{h})$ is injective. If $G$ is simply connected, then it is bijective: that is, every map $\mathfrak{g} \to \mathfrak{h}$ of Lie algebras exponentiates to a unique map $G \to H$ of Lie groups.

Taking $H = GL_n(\mathbb{R})$ or $GL_n(\mathbb{C})$ it follows that a simply connected Lie group $G$ and its Lie algebra $\mathfrak{g}$ have the same (finite-dimensional) representation theory (over $\mathbb{R}$ or over $\mathbb{C}$).

This is standard Lie theory and you should be able to find it in any good book on Lie groups and/or representation theory; for example it's stated and proven right after Exercise 8.42 in Fulton and Harris' Representation Theory: a First Course.

It's worth mentioning that for applications to quantum mechanics you're often happy to recover just a projective representation (since you still get a genuine action on states regarded as points in the projective space), and then we have the following:

Proposition 2: If $G$ is a connected Lie group and $\rho : \mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$ is an irreducible complex representation of its Lie algebra $\mathfrak{g}$, then it always exponentiates to a projective representation $G \to PGL_n(\mathbb{C})$.

Sketch. Using Proposition 1 we recover an irreducible representation of the universal cover $\widetilde{G} \to GL_n(\mathbb{C})$. It's another standard Lie theory fact that the kernel of the covering map $\widetilde{G} \to G$ (which can be identified with the fundamental group $\pi_1(G)$) is a discrete central subgroup $Z$ of $\widetilde{G}$. Now by Schur's lemma $Z$ acts by a scalar, so the action of any two lifts of $g \in G$ to $\widetilde{G}$ differ by the action of an element of $Z$ and hence by a scalar, which exactly says that we get a projective representation $G \to PGL_n(\mathbb{C})$. $\Box$

For example, the spin representations are projective representations of $SO(N)$ where the nontrivial element in the kernel of the map $\text{Spin}(N) \to SO(N)$ acts by $-1$.

  • $\begingroup$ Thanks for the answer @QiaochuYuan ! I'm still puzzled by something, though. ${\rm SU}(N)$ is compact and connected, the exponential is surjective and hence cannot be injective. In that setting, why does being simply connected solves the ambiguity? I mean, let ${\bar D}$ be a ${\frak{su}}(N)$ irrep. We want to define a ${\rm SU}(N)$ irrep $D$. Let $g\in {\rm SU}(N)$. There is still $X\neq X'$ with $g = \exp(X)=\exp(X')$ and so we can either define $D(g) = \exp \bar D(X)$ or $D(g)=\exp \bar D(X')$. How ${\rm SU}(N)$ being simply connected makes $D(g)$ well defined? $\endgroup$ – Gold Oct 9 '20 at 13:30
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    $\begingroup$ @user1620696: that’s exactly what Proposition 1 is implying. This is not at all obvious! You can take a look at the proof in Fulton and Harris. $\endgroup$ – Qiaochu Yuan Oct 9 '20 at 17:03

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