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I would appreciate your insight on how to prove that a complex matrix group is a Lie Group.

Let $G =\left\{ U(\alpha, \beta) = \left( \begin{array}{cc} \alpha & \beta \\ \beta^* & \alpha* \end{array} \right)\; | \; det \, U(\alpha, \beta) = 1 \right\}$ be a group (I am able to show this).

To prove that this is in fact a Lie Group, I need to prove that it is indeed a manifold, and then show that the group operations multiplication and inverse are smooth.

First, I construct $\phi : G \rightarrow \mathbb R^4 - \{x, y, z, t \; | \; x^2 + y^2 = z^2 + t^2\}$ (open set of $\mathbb R^4$) defined by $\phi \left(U(\alpha, \beta)\right) = (Re(\alpha), Im(\alpha), Re(\beta), Im(\beta))$. $\phi$ is a bijective map. There is only one transition chart which is the identity, which is smooth. Is that enough to show that G is a manifold?

The elements of the product of two matrices being polynomials of the elements of the matrices, the multiplication map is smooth. Same for the inverse since the matrices have determinant 1.

Does this prove that $G$ is a Lie Group or am I missing something?

Thank you and have a good day!

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  • $\begingroup$ What are $\alpha^*$ and $\beta^*$? Do you mean $\overline\alpha$ and $\overline\beta$? $\endgroup$ Commented Oct 21, 2018 at 14:43
  • $\begingroup$ Yes sorry, the complex conjugates of $\alpha$ and $\beta$ $\endgroup$ Commented Oct 21, 2018 at 14:43
  • $\begingroup$ The proposed map $\phi$ is not a bijection: The condition on $U(\alpha, \beta)$ in the definition of $G$ imposes that any point $(x, y, z, t)$ in the image of $\phi$ satisfies $x^2 + y^2 - z^2 - t^2 = 1$. $\endgroup$ Commented Oct 21, 2018 at 14:58
  • $\begingroup$ Oh right, I meant $=$ not $\neq$, thank you $\endgroup$ Commented Oct 21, 2018 at 15:22

2 Answers 2

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Hint:A closed subgroup of a Lie group is a Lie group, so you just have to show that it is closed, if you consider a sequence $A_n\in G$, such that $A=lim_nA_n, det(A)=1$ since $det$ is continuous and if $lim_n\alpha_n=\alpha, lim_n\alpha_n^*=\alpha^*$.

Another way to see tis is to consider the space of matrices $L=\{\pmatrix{\alpha &\beta\cr \beta^*&\alpha^*}\}$, and to show the map $det:M\rightarrow\mathbb{C}$ is a submersion on a neighborhood of $G$, this shows that $G$ is a submanifold.

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  • $\begingroup$ Thank you, nevertheless I am supposed to show it from the definitions only, not assuming general facts about Lie Groups (I have just started learning about them so I don't know the relevants theorems for now). $\endgroup$ Commented Oct 21, 2018 at 14:48
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Hint It's easier not to construct charts, and instead deduce smoothness from some ambient set in which you know that $G$ is embedded.

Construct a smooth map $F$ from $GL(2, \Bbb C)$ for which $G$ is a level set $F^{-1}(c)$. If you know that by Cartan's Theorem a closed subgroup of a Lie group is a Lie subgroup, then you're done.

If you don't know that, pick such an $F$ such that $c$ is a regular value, which establishes that $G = F^{-1}(c)$ is a smooth embedded manifold. Then, like you suggest, the multiplication and inversion maps are smooth because they have rational components.

Additional hint Take $H$ to be the set of Hermitian matrices in $A \in M(2, \Bbb C)$, that is, those satisfying $A^* = A$, and define $F : GL(2, \Bbb C) \to H$ by $A \mapsto A^* \Sigma A$, where $\Sigma = \pmatrix{1&\\&-1}$. Then, $G = F^{-1}(\Sigma)$.

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