# Why can we always lift representations of the Lie algebra $\mathfrak{su}(N)$ to representations of the Lie group ${\rm SU}(N)$?

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?

## 1 Answer

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 $$-1$$.

• 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? – Gold Oct 9 '20 at 13:30
• @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. – Qiaochu Yuan Oct 9 '20 at 17:03