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I understand that the Metaplectic group is as a real Lie group that is the unique double cover of the symplectic group (specifically the split real form, in case it is unclear).

However whenever I read about the Metaplectic group, the field $\mathbb{C}$ is always excluded as a possibility. Can one not simply take the generators, allow complex combinations of generators, and see what falls out? Why is the metaplectic group not defined when the underlying field is $\mathbb{C}$?

Craig
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    See here. The symplectic group $Sp_{2n}(\Bbb C)$ is simply connected, so a cover will be the group itself. – Dietrich Burde Jun 26 '23 at 18:35
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    Yes that is clear. I guess the heart of the question is what "happens" to the Metaplectic group under complexification then? Does it somehow "fall back down" to being a (non-projective) representation of the Symplectic group? – Craig Jun 26 '23 at 18:54
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    I've always had a hard time understanding what complexification of a group "actually is" (and I'm not the only one, cf. e.g. https://math.stackexchange.com/q/1298547/96384) but user Callum's answer here https://math.stackexchange.com/a/4333326/96384 is the best intro I've read. I think it implies that the metaplectic group (having no faithful matrix rep) just "does not have" a complexification in one sense of the word. But it might still have one in the sense of "universal complexification". Maybe try going through the steps in Callum's answer about that, I'd be eager to see what comes out! – Torsten Schoeneberg Jun 26 '23 at 19:28
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    (Although I strongly suspect what comes out is trivial as per what @DietrichBurde suggests. Might still be good to see how to arrive at that simple result through the technicalities.) – Torsten Schoeneberg Jun 26 '23 at 19:29
  • Thank you Torsten. I'll have to do some reading to understand the meaning here of representation maps factorizing, I hadn't seen that before. I agree the question is more interesting in general, however for my particular purposes, the relevant question is "If after fixing a basis in my real Lie algebra $\mathfrak{mp}(n,\mathbb{R})$, I allow complex combinations of my basis, does the same exponential map from the real algebra to the real group, send my "complexified algebra" to any well defined group?" I'll keep you updated if I make any progress in understanding this. – Craig Jun 26 '23 at 20:32
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    @Craig the problem with that last suggestion is that to extend the exponential map to a complexification you really need to implicitly assume the original group is linear. In that case, you can use the power series definition of $\operatorname{exp}$. How else could you define it after all? For a nonlinear group like the metaplectic group you can't do this so it has no complexification (apart from a universal one which is the complex symplectic group) – Callum Jun 26 '23 at 22:15
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    Apologies, I am a physicst, do you know where I can read more specifically about the relationship between the exponential map and group linearity? If the exponential map doesn't work for the metaplectic group, how do we go from algebra elements to group elements even in the real case? Thank you – Craig Jun 27 '23 at 01:00
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    To be clear, there is an exponential map for the metaplectic group it just doesn't have a power series representation. It's important to note the exponential map only makes sense once we've picked a group for it to be in and different groups with the same Lie algebra have different exponential maps. Working with matrix groups hides this as once you've picked a faithful representation of the Lie algebra the power series definition of exp picks out a group. – Callum Jun 27 '23 at 06:50
  • If you can recommend readings on this it would be most appreciated. Thank you again. – Craig Jun 27 '23 at 07:11

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Was going to be a bit of a long comment so I've turned it into an answer. I am actually not sure of a good reference for this topic. Knapp's Lie Groups: Beyond an Introduction goes into real groups and covering groups a bit but I don't have a copy to check whether he discuss the metaplectic group or complexification in this manner.

The key here is that complexification is really a linear thing. We complexify real vector spaces and only from that can we obtain a way to complexify a group.

Throughout I will use the following definitions: a complexification of a real Lie group $G$ is a complex Lie group $G^\mathbb{C}$ such that $G \leq G^\mathbb{C}$ and the Lie algebra $\mathfrak{g}$ of $G$ is a real form of the Lie algebra $\mathfrak{g}^\mathbb{C}$. A universal complexification is a complex Lie group $G'$ equipped with a homomorphism $\phi: G \to G'$ with the universal property that any homomorphism from $G$ to a complex Lie group $H$, $f:G\to H$ there is a unique complex analytic homomorphism $F:G'\to H$ such that $f = F \circ \phi$. It should be clear that if a complexification exists then it is universal ($G^\mathbb{C} = G'$) with $\phi$ the inclusion map.

Firstly, the complexification. If $G \leq GL(n,\mathbb{R})$ then as you suggest it is not hard to find a complexification by simply restricting the exponential map $\exp:\mathfrak{gl}(n,\mathbb{C})\to GL(n,\mathbb{C})$ to $\mathfrak{g}^\mathbb{C}$ which will then extend the exponential map $\exp:\mathfrak{g}\to G$. In essence we are making use of the well conceived notion of complexification for $GL(n,\mathbb{R})$. Since you can define $\exp(X) = I + X + \frac{1}{2}X^2 + \dotsm$, you can even directly compute the exponential map on complex linear combinations of our real elements up to BCH formula computations etc. and certainly $\exp(iX) = I + iX - \frac{1}{2}X^2 + \dotsm$ makes sense.

It is crucial at this point to understand that a Lie algebra does not just have one exponential map. It has one for every Lie group it has and they are not the same thing. For example, $\exp:\mathfrak{sl}(n,\mathbb{R})\to SL(n,\mathbb{R})$ and $\exp:\mathfrak{sl}(n,\mathbb{R})\to PSL(n,\mathbb{R})$ are different maps.

Secondly, the universal complexification. As I discuss here you can construct $G'$ by first considering the homomorphism between the real and complex simply connected groups $\tilde{G}$ and $\tilde{G}_\mathbb{C}$, $\Phi:\tilde{G} \to \tilde{G}_\mathbb{C}$. Note the latter doesn't have to be the complexification of the former but we can still find such a $\Phi$ as the unique one whose differential is the inclusion map $\mathfrak{g}\hookrightarrow \mathfrak{g}^\mathbb{C}$. Simultaneously, we consider the projection $\pi: \tilde{G} \to G$ and define $G' := \tilde{G}_\mathbb{C}/N$ where $N$ is the smallest normal subgroup containing $\Phi(\ker \pi)$.

A few observations:

  • This always produces a complex Lie group even when there is no complexification
  • It can happen that the Lie algebra of $G'$ is not $\mathfrak{g}^\mathbb{C}$ but some quotient of it. This is not an issue in the semisimple case since $N$ will be discrete.
  • If $G = \tilde{G}$ then $G' = \tilde{G}_\mathbb{C}$ since $\pi$ will be trivial.

As you can see this is a bit more of an abstract idea than the usual complexification but it always works.

Callum
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