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I have proved that $U(n)$, which is a group of unitary matrices, is a smooth manifold of real Dimension $n^2$. Now I am trying to show that $SU(n)$, which is a group of unitary matrices with determinant 1, is a smooth submanifold of $U(n)$ of real dimension $n^2-1$. However I am stuck at a step:

The idea is to use the theorem: Suppose $f:N^n \to M^m$ is a smooth map between smooth manifolds $N,M$ of dimension $n,m$ respectively. Then preimage of any regular value $y$ of $f$ is a regular submanifold of $M$ and is of dimension $n-m$. Here, $y$ is a regular value means that the rank of $f$ at every point in the preimage of $y$ has rank exactly $m$.

To use this Theorem, I considered the determinant map

$$ det: U(n) \to S^1:=\{z \in \mathbb{C}:|z|=1\}. $$ Now I want to show that $1 \in S^1$ is a regular value of the map $det.$ That is, I have to show that derivative of $det$ at an arbitrary $g\in det^{-1}(1)=SU(n)$ has rank $1$. For an arbitrary $Y \in Mat_{n \times n}\mathbb{C}$, I get $$D_gdet(Y)=\lim_{t \to 0}\frac{det(g+tY)-det(g)}{t}=\lim_{t \to 0}\frac{det(g+tY)-1}{t}$$

I don`t know how should I proceed from here. How can I simplify the RHS?

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    Have you tried starting at the special case $g=1$ ? – Max Mar 14 '16 at 20:20
  • You can note that $D_g\det(Y)=D_I\det\left(g^{-1}Y\right)$; so that, without loss the generality, you can compute $D_I\det(Y)$ and you use the Taylor series formula fort $\det(Y)$ near to $I$, the identity matrix. – Armando j18eos Mar 15 '16 at 18:44

2 Answers2

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You need to calculate the derivative of $\det$, which is

$$D_g \det (Y) = \det (g) \text{tr}(g^{-1}Y) = \text{tr} (g^{-1} Y) \in i\mathbb R = T_1\mathbb S^1 $$

as $\det (g) = 1$. Note that it suffices to find, for all $g\in \det^{-1}(1)$, $Y\in T_gU(n)$ so that $D_g\det(Y) \neq 0$. Note that

$$T_gU(n) = g\cdot u(n),$$

where $u(n)$ is the Lie algebra of $U(n)$ given by

$$u(n) = \{ A \in M_n(\mathbb C) : A + A^* = 0\}$$

Then $iI_n \in u(n)$ and $Y = g(iI_n)$ satisfies

$$D_g \det (Y) = \text{tr} (g^{-1} g(iI_n)) = ni\neq 0.$$

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I don't know if you're in the context of Lie Groups. In this case you just need to invoke the Closed Subgroup Theorem which states that every closed subgroup of a Lie Group is a Lie Group, which also means by definition that is a submanifold.

$SU(n)$ is a closed subgroup of $U(n)$ hence a submanifold. To see that is closed just consider the function determinant.

Dac0
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  • Is there a way to know the dimension of $SU(n)$ using the above or similar theorem? –  Mar 14 '16 at 22:13