# Clarification of relation between $SL(2,\mathbb{R})$ and $Sp(2,\mathbb{R})$

I have been reading up about the homomorphism between $$SO^+(2,1)$$ (i.e. the proper-orthochronus Lorentz group in 2+1 dimensions), and $$SL(2,\mathbb{R})$$. As part of this, I was further introduced to the symplectic group $$Sp(2,\mathbb{R})$$, and this is where my confusion arises. From what I've read, it is claimed that $$Sp(2,\mathbb{R})$$ and $$SL(2,\mathbb{R})$$ are isomorphic, i.e. $$Sp(2,\mathbb{R})\cong SL(2,\mathbb{R})$$, however, as fair as I can tell, from their group structures, they seem to be not just isomorphic, but identical. By this, I mean that the isomorphism between them seems to be trivial.

Am I missing something here, or is it just a quirk of the $$n=2$$ case, that the two groups are trivially related?

If I am correct, then they essentially seem to be the same group in this case (rather than there existing a one-to-one mapping between them), and so why do people even distinguish them?

If I am incorrect, can someone please enlighten me, and also detail the actual isomorphism?

To caveat this question, I am approaching this from a physicist's perspective, so apologies for any lack of mathematical rigour, and/or discrepancies in notation/conventions.

• You're right, this is a quirk about the case $n=2$. In higher dimensions, there are matrices with determinant 1 which are not symplectomorphisms (try the matrix with 1's at (1,1),(4,2),(2,3),(3,4) and zeroes elsewhere). Sep 16, 2020 at 16:30
• @MichaelGintz Ah ok. So are they simply the same group in $n=2$ then? Why do people even bother dressing it up as an isomorphism, since the mapping between the two seems to be the identity?
– Will
Sep 16, 2020 at 16:51
• Related (wrong title) $Sp(2n, \Bbb R) = SL(2n, \Bbb R)$ Sep 16, 2020 at 16:54

## 1 Answer

Yes, you're right that if you think of $$SL_2(\mathbb{R})$$ as the particular set of matrices in $$GL_2(\mathbb{R})$$ of determinant $$1$$ and if you think of $$Sp_2(\mathbb{R})$$ as the particular set of matrices in $$GL_2(\mathbb{R})$$ preserving the standard symplectic form on $$\mathbb{R}^2$$ then they are literally the same matrices.

There are (at least) three additional issues here.

1. When a mathematician says "$$SL_2(\mathbb{R})$$" (or any other mathematician object) they aren't talking about a particular set of matrices sitting inside $$M_2(\mathbb{R})$$. Generally they're talking, implicitly, about the isomorphism class of all Lie groups isomorphic to $$SL_2(\mathbb{R})$$, although perhaps they also want to equip it with its defining representation on $$\mathbb{R}^2$$. The point of doing this is to have the flexibility to work with mathematical objects without needing to pick a particular representation of them. For example, when I talk about the cyclic group $$C_2$$ of order $$2$$ I am giving myself the freedom to not tie it to a concrete description such as "the group $$\{ \text{even}, \text{odd} \}$$ under addition $$\bmod 2$$" or "the group $$\{ 1, -1 \}$$ under multiplication" and so forth.

2. $$SL_2(\mathbb{R})$$ is canonically defined as a subgroup of $$GL_2(\mathbb{R})$$ because the determinant is canonically defined, but defining $$Sp_2(\mathbb{R})$$ requires a choice of symplectic form. Now, it happens that in dimension $$2$$ there's a unique symplectic form up to scale, so only one subgroup of $$GL_2(\mathbb{R})$$ occurs this way, but this is sort of a low-dimensional coincidence and in general $$Sp_{2n}(\mathbb{R})$$ is not a canonically defined subgroup of $$GL_{2n}(\mathbb{R})$$. It's a convention that we have a "standard" symplectic form in mind on $$\mathbb{R}^{2n}$$ (and I can think of at least two possible conventions) and $$Sp_{2n}(\mathbb{R})$$ as a subgroup (although not as a group-up-to-isomorphism) in general depends on this choice.

3. We give them different names because they're part of two different sequences of groups that are constructed in two different ways, that just happen to coincide near the beginning. There are several such exceptional isomorphisms and they don't stop us from using multiple names for things. As a simpler example, the third Fibonacci number is $$F_3 = 2$$ but that doesn't mean we should decide to only ever call this number either "$$2$$" or "$$F_3$$," we can and do do both.

• +1, I disagree a bit on point one, I think $\text{SL}_2(\mathbb{R})$ really means the subgroup of $M_2(\mathbb{R})$ and if you wanted to mean (any representative of the class of Lie groups abstractly isomorphic to this) it would be better to say let $V$ be a two-dimensional real vector space and $G = SL(V)$. (whereas $C_2$ basically means what you said.) Sep 16, 2020 at 16:55
• That might be a good convention in some contexts but I don't think it's necessary or standard. When people say "the symmetric group $S_n$" it's usually clear they're referring to an isomorphism class of group without them needing to say "let $X$ be a set with $n$ elements and consider $G = \text{Aut}(X)$." Plus it would be very wordy. Sep 16, 2020 at 16:58
• Thanks for your detailed answer. So, in essence, the only way that they differ for $n=2$ (if you consider the group elements as matrices), is that one has to, in principle, choose a symplectic form in order to define $Sp_2(\mathbb{R})$? I thought, though, that $SL_2(\mathbb{R})$ implies this symplectic form anyway, i.e. the Levi-Civita tensor is an invariant under $SL_2(\mathbb{R})$ transformations, which implies that the elements of $SL_2(\mathbb{R})$ preserve the symplectic form too?
– Will
Sep 16, 2020 at 17:02
• @Will: yes, again, in dimension $2$ there's a unique symplectic form up to scale and its symplectic group is $SL_2(\mathbb{R})$, because in dimension $2$ symplectic forms are the same as (oriented) volume forms. So up to scale there is practically no difference, and $SL_2(\mathbb{R})$ preserves every symplectic form on $\mathbb{R}^2$. But very strictly speaking the definition of a symplectic group requires a choice of symplectic form, full stop, not just one up to scale. Sep 16, 2020 at 17:06
• @QiaochuYuan I am reminded of this question math.stackexchange.com/questions/181464/… There seems to be a hard split between users who think $\mathbb{C}$ means some abstract algebraically closed field containing $\mathbb{R}$ and those who think it means formal expressions $a + bi$; the former group thinks there is no canonical square root of $-1$ in $\mathbb{C}$ and the latter thinks there is an obvious one. Very similarly, I'd argue that $SL_2(\mathbb{R})$ has a canonical maximal compact and you would say it doesn't I guess. Sep 18, 2020 at 19:19