Exponentiating a representation of a semi-simple Lie algebra

I consider a representation of a semi-simple Lie algebra $$\mathfrak{g}$$ (specifically, the symplectic and orthogonal Lie algebras $$\mathfrak{sp}(2N)$$ and $$\mathfrak{so}(2N)$$) as anti-Hermitian operators acting on some Hilbert space $$\mathcal{H}$$, i.e., $$\rho(K)\rho(K')-\rho(K')\rho(K)=\rho([K,K'])$$ and $$\rho(K)^\dagger=-\rho(K)$$ for $$\rho: \mathfrak{g}\to \mathrm{Lin}(\mathcal{H})$$.

My understanding is that if we now exponentiate $$e^{\rho(K)}$$ and take arbitrary products of these operators, we will get a unitary representation of a group, whose Lie algebra is $$\mathfrak{g}$$. Specifically, if $$\mathfrak{g}$$ is the symplectic algebra, the resulting group will be symplectic group or one of its $$n$$-fold covers (potentially even its universal cover with $$n=\infty$$). If $$\mathfrak{g}$$ is the orthogonal algebra $$\mathfrak{so}(2N)$$ with $$N>1$$, the resulting group would be either $$SO(2N)$$ or its double cover $$Spin(2N)$$, which is also its universal cover.

My questions: Are there requirements that I am missing to make this true? And under the conditions where this is true, do you know a standard reference that shows this or closely related statement?

Thank you so much.

• Is your Lie algebra $\mathfrak g$ considered as a real Lie algebra? Is $\mathcal H$ finite-dimensional? Commented Jul 9 at 16:26
• Yes, I consider the case $\mathfrak{g}$ being real, but would be curious to understand the general case. $\mathcal{H}$ could be both finite or infinite dimensional. Specifically, I'm interested in the cases where $\mathfrak{g}=\mathfrak{sp}(2N,\mathbb{R})$ with infinite dimensional $\mathcal{H}$ and $\mathfrak{g}=\mathfrak{so}(2N,\mathbb{R})$ with finite dimensional $\mathcal{H}$. My hope was that there is some general understanding...
– LFH
Commented Jul 9 at 22:24

First I suppose that $$\mathcal H$$ is finite-dimensional. Lie’s third theorem implies that a Lie homomorphism $$\mathfrak g \to \mathfrak u(\mathcal H)$$ integrates to a group homomorphism $$\tilde G \to U(\mathcal H)$$ where $$\tilde G$$ is the connected simply connected Lie group with Lie algebra $$\mathfrak g$$. Not only that, but if $$\mathfrak g$$ is semisimple, then after complexifying, we find that this representation actually factors through $$G_{\mathbb C} \to GL(\mathcal H)$$, so the representation of $$\tilde G$$ factors through the image of $$\tilde G \to G_{\mathbb C}$$. (For $$\mathfrak{so}$$ this is the spin group, for $$\mathfrak{sp}$$ this is Sp.) Lie’s third theorem can be found in any text on Lie theory.

If $$\mathcal H$$ is infinite-dimensional this is more subtle. Usually $$\mathfrak g$$ only acts by densely defined operators. If $$K$$ is the maximal compact of $$G$$, then Harish-Chandra worked out how to integrate when $$K$$ acts locally finitely on the so-called smooth vectors. See eg papers of Vogan for this.

A famous example of such a representation for $$\mathfrak{sp}$$ is the Weil representation. Maybe this is the representation you seek.

• Thank you so much. I'm a bit confused about your first statement. If we consider the example of the Lie algebra $\mathfrak{g}=i \mathbb{R}$, i.e, imaginary numbers (acting on a one-dimensional complex Hilbert space), it would exponentiate to $U(1)$, which is not simply connected. Essentially, my intuition was that a given anti-Hermitian representation of the Lie algebra should exponentiate to some representation of a Lie group, whose Lie algebra is $\mathfrak{g}$, but we cannot say a priori which one.
– LFH
Commented Jul 10 at 13:13
• There is a unique connected simply connected such group. If $\mathfrak g = i\mathbb R$ then $\tilde G = i\mathbb R$ and that’s it. It’s a perfectly good Lie group, but it’s not compact. You can write down when it descends to $U(1)$ in terms of the Jordan form of the differential. Commented Jul 10 at 13:16
• Thank you! Yes, I agree that there is such a unique group $\tilde{G}$. However, if I have a specific representation $\mathfrak{g}\to \mathfrak{u}(\mathcal{G})$ of the algebra and I then apply the exponential map (to the matrices of the representation), I may not get a representation of $\tilde{G}$, e.g., if we represent $\mathfrak{g}$ by $i \mathbb{R}$ (1x1 matrices with $\mathcal{H}$ being 1-dimensional), exponentiation clearly gives $e^{i \mathbb{R}}$, which is the unit circle in the complex plane. Once I fix a representation of $\mathfrak{g}$, the exp map may not yield a rep of $\tilde{G}$.
– LFH
Commented Jul 11 at 0:03
• "exponentiation clearly gives $e^{i\mathbb R}$, which is the unit circle in the complex plane." At the level of exponentiating a matrix, this is not correct. Commented Jul 11 at 1:16
• I'm confused? Probably that's where I don't follow the argument. Thank you so much for all the clarification you are providing. If I have a 1x1 matrix with imaginary entries (here a single entry) and then apply the exponential map defined as $\exp(M)=\sum^\infty_{n=0}M^n/(n!)$, I get a 1x1 matrix whose complex entry lies on the unit circle, so here I would say that the representation of the Lie algebra picks a specific group that is represented by its exponential (here: $U(1)$).
– LFH
Commented Jul 11 at 5:05