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During a lecture of a Lie Algebras yesterday, the professor of the class stated the following fact without proof

$O_{2n}(\mathbb{R}) \cap GL_{n}(\mathbb{C})=U(n)$

Note that we are viewing $GL_{n}(\mathbb{C})$ as elements of $GL_{2n}(\mathbb{R})$

I was wondering if there was a quick way to see this ?

I can see why this is so at $n=1$, where we need to show that $O_{2}(\mathbb{R}) \cap GL_{1}(\mathbb{C})=U(1)$

We can identify elements of $GL_{1}(\mathbb{C})$ with $2 \times 2$ matrices of the form

$\begin{bmatrix} x&-y\\ y&x\\ \end{bmatrix}$ where $x,y \in \mathbb{R}$ and whose determinant $x^{2}+y^{2}$ in nonzero.

We can identify $U(1)$ with the rotation group of the plane $SO(2)$. For a matrix $A \in O_{2}(\mathbb{R}) \cap GL_{1}(\mathbb{C})$, we that $det(A)=x^{2}+y^{2}$ is either $1$ or $-1$ and the sum of squares implies that the determinant of $A$ is $1$ and hence $A$ must be in $SO(2)$.

I realize now that this has to do with the 2 out of 3 property

https://en.wikipedia.org/wiki/Unitary_group#2-out-of-3_property

Thanks in advance.

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Your given embedding $\mathit{GL}_n(\Bbb C) \hookrightarrow\mathit{GL}_{2n}(\Bbb R)$ replaces each complex entry $$a+ib$$ in your given $n\times n$ complex matrix with the $2\times 2$ block $$\begin{matrix} a & -b \\ b & a\end{matrix}$$ of real entries to form a $2n\times 2n$ real matrix.

Proving the desired intersection is easy if you characterize unitary matrices as those (complex) matrices $M$ such that $M^{-1} = M^*$, where $M^*$ is the conjugate transpose, and orthogonal matrices as those (real) matrices $A$ such that $A^{-1} = A^T$. Your desired statement is then a formal consequence of the following two facts:

  1. The embedding $\phi: \textit{GL}_n(\Bbb C) \hookrightarrow \textit{GL}_{2n}(\Bbb R) $ is a homomorphism of groups, i.e. it preserves matrix multiplication $\phi(M_1M_2) = \phi(M_1)\phi(M_2)$. (In particular, the image of an inverse is the inverse of the image.)
  2. The embedding $\phi$ converts conjugate transposes to transposes:for any $M\in \textit{GL}_n(\Bbb C)$, we have $\phi(M^*) = \phi(M)^T$.

With these two facts we see that $M^{-1} = M^*$ if and only if $\phi(M)^{-1} = \phi(M)^T$, as desired.

The first fact is necessary to even view $\textit{GL}_n(\Bbb C)$ as elements of $\textit{GL}_{2n}(\Bbb R)$, and is straightforward to verify, although the notation might be a bit messy.

The second fact is also immediate, since taking the transpose of a $2n\times 2n$ real matrix in the image of $\phi$ consists of externally transposing the $2\times 2$ blocks with one another and then internally replacing each block $\begin{matrix} a & -b \\ b & a\end{matrix}$ with the block $\begin{matrix} a & b \\ -b & a\end{matrix}$. This corresponds precisely to taking the conjugate transpose of the original matrix, as desired.

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  • $\begingroup$ Thanks friend, this is what I was looking for. $\endgroup$ – user135520 Aug 27 '15 at 21:58

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