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Consider following Lie Group: $$ \text{Sp}(2n,\mathbb{C})=\{g\in\text{Mat}_{2n}(\mathbb{C})\mid J=g^TJg\}\quad\ where\quad J=\begin{pmatrix} 0 & 1_n \\ -1_n & 0 \end{pmatrix} $$ And the corresponding Lie Algebra: $$ \text{sp}(2n,\mathbb{C})=\{g\in\text{Mat}_{2n}(\mathbb{C})\mid g^TJ+Jg=0\} $$

Are there any basic proofs that $\text{Sp}(2n,\mathbb{C})$ is a Lie Group and that $\text{sp}(2n,\mathbb{C})$ is the corresponding Lie Algebra without using submersions (seen here: Why is $Sp(2m)$ as regular set of $f(A)=A^tJA-J$, and, hence a Lie group.)? Can I proove the first statement by showing that $\text{Sp}(2n,\mathbb{C})$ is a subgroup of $\text{GL}(n,\mathbb{C})$?

Thanks!

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  • $\begingroup$ Subgroup is not enough. You also need that it is a submanifold. $\endgroup$ – Daniel Fischer Jun 26 '13 at 9:30
  • $\begingroup$ If you're willing to accept the Closed Subgroup Theorem, then I think you can find a more basic proof. However, the Closed Subgroup Theorem is somewhat involved, by no means trivial. $\endgroup$ – Jesse Madnick Jun 26 '13 at 9:36
  • $\begingroup$ On the Lie group part. You can identify $Sp(2n,\mathbb C)$ with a closed subset of $\mathbb C^{4n^2}$ using the usual distance between matrices as metric. The idea is that each matrix in $Sp(2n,\mathbb C)$ is a long vector with $4n^2$ components satisfying the complicated equality in the definition of $Sp(2n,\mathbb C)$ itself. Once you have done this, you should prove that the composition and inverse maps are smooth. The composition produces polynomials, while the inverse map can be studied with the Cramer rule. $\endgroup$ – Avitus Jun 26 '13 at 9:59
  • $\begingroup$ For the first Part i tried the following thing: Let $G \subset\mathbb{C}^{4n^2}$ be a closed subset of $\mathbb{C}^{4n^2}$. $$\vec{j}=\vec{v}^T\vec{j}\vec{v}\quad\text{is valid}\quad\forall\vec{v}\in\mathbb{C}^{4n^2}\quad(\text{The equation satisfying}\quad\text{Sp}(2n,\mathbb{C}))$$ Defining then the map: $$ f(\vec{v})=\vec{v}^T\vec{j}\vec{v}-\vec{j} $$ Is this the right function to show composition and inverse map? What is then the purpose of the metric? Hmm I think Im getting it wrong.. $\endgroup$ – user83926 Jun 26 '13 at 13:25
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On the Lie algebra part. By definition, $sp(2n,\mathbb C)$ is the real vector space of complex matrices

$$sp(2n,\mathbb C)=\{ A\in Mat_{2n}(\mathbb C): \exp(tA)\in Sp(2n,\mathbb C), \forall t\in \mathbb R \}.$$

$\exp(tA)$ denotes the exponential of the matrix $tA$. By definition of $Sp(2n,\mathbb C), $ we have that

$$J=\exp(tA)^{t}J\exp(tA)=\exp(tA^{t})J\exp(tA), (*)$$

where the second equality can be proved using the definition of the exponential of a matrix.

Using ($\exp(0)=1$)

$$\frac{d\exp(tA)}{dt}|_{t=0}:=\lim_{t\rightarrow 0}\frac{\exp(tA)-1}{t}= \lim_{t\rightarrow 0}\frac{(1+tA+\frac{1}{2!}t^2A^2+\dots)-1}{t}=A$$

we arrive at

$$0=(\text{using (*)})=\frac{d}{dt}(J-\exp(tA)^{t}J\exp(tA))|_{t=0}=0-A^{t}J+JA,$$

i.e. $A^{t}J+JA=0$, as $J$ is independent of $t$.

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  • $\begingroup$ You are welcome! $\endgroup$ – Avitus Jun 26 '13 at 12:26

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