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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.

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    $\begingroup$ Is your Lie algebra $\mathfrak g$ considered as a real Lie algebra? Is $\mathcal H$ finite-dimensional? $\endgroup$
    – Joshua Mundinger
    Commented Jul 9 at 16:26
  • $\begingroup$ 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... $\endgroup$
    – LFH
    Commented Jul 9 at 22:24

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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.

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  • $\begingroup$ 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. $\endgroup$
    – LFH
    Commented Jul 10 at 13:13
  • $\begingroup$ 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. $\endgroup$
    – Joshua Mundinger
    Commented Jul 10 at 13:16
  • $\begingroup$ 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 $\tilde{G}$. $\endgroup$
    – LFH
    Commented Jul 11 at 0:03
  • $\begingroup$ "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. $\endgroup$
    – Joshua Mundinger
    Commented Jul 11 at 1:16
  • $\begingroup$ 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: $U(1)$). $\endgroup$
    – LFH
    Commented Jul 11 at 5:05

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