# The dimension of $SU(n)$

$SU(n)$ denotes the special unitary group. I know its dimension should be $n^2-1$. However, I am trying to prove it and get a wrong result. I have no idea what is wrong with my proof. Therefore, I am wondering if someone could point the mistake out. Thank you! The following is my proof.

I consder the determinant function $\det:U(n)\to\mathbb{C}$, where $U(n)$ denotes the unitary group. $\det$ should be a Lie group homomorphism. Hence, it is a smooth map with constant rank. Then, $SU(n)=\det^{-1}(1)$, which actually shows that $SU(n)$ is an embedded Lie subgroup of $U(n)$.

I have already known that $U(n)$ is $n^2$-dimensional. It suffices to figure out the rank of $\det$, since the codimension of $SU(n)$ is the rank of $\det$.

It suffices to check the rank at $I_n$ which denotes the identity matrix, since $\det$ has a constant rank. Pick $B\in T_{I_n}U(n)=M(n,\mathbb{C})$, where $M(n,\mathbb{C})$ denotes the set of all matrices whose entries are complex.

Therefore, the differential of $\det$ is a map $d(\det)_{I_n}:M(n,\mathbb{c})\to\mathbb{C}$. Consider the curve $$\gamma(t)=I_n+tB$$. Clearly, $\gamma'(0)=B$, $\gamma(0)=I_n$.

Then, I compute \begin{align*} d(\det)_{I_n}(B) &=(\det\circ\gamma)'(0)\\ &=\frac{d}{dt}|_{t=0}\det(I_n+tB)\\ &=tr(B) \end{align*}

Now, I think I can prove the image of $d(\det)_{I_n}$ is $\mathbb{C}$, since given any $c\in\mathbb{C}$ I take the matrix $B$ as $\frac{c}{n}I_n$.

Therefore, I believe that the rank of $\det$ is 2, which is the dimension of $\mathbb{C}$. So I conclude that the dimension of $SU(n)$ is $n^2-2$.

I know I must make a silly mistake. Could someone point it out? Thank you very much!

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The determinant on $U(n)$ has rank at most rank $1$ since it takes values in the circle $S^1\subset\Bbb C^{\times}$, a one dimensional manifold. –  Olivier Bégassat May 16 '13 at 14:25
Alternatively, $T_I U(n) =\mathfrak u(n)$ is the vector space of skew-hermitian matrices, so the diagonal entries are all pure-imaginary. Thus, to say the trace is $0$ is one real condition. In your argument, $B$ cannot be any complex matrix. It must belong to $\mathfrak u(n)$. –  Ted Shifrin May 16 '13 at 14:38
@OlivierBégassat Could you point out the mistake I made in my proof? I know there must be lots of ways to calculate the dimension. Besides, I know $|\det A|=1$ holds for all $A\in U(n)$. Anyway, thank your for your comment. –  Y. Fan May 16 '13 at 14:44
@TedShifrin Did you mention the Lie Algebra? I have not learnt it yet. Here is my thought. $U(n)$ is an embedded submanifold (or regular submanifold) in $M(n,\mathbb{C})$ and a vector space. Therefore, its tangent spaces are cannonically identified with $M(n,\mathbb{C})$. Therefore, I wrote $T_IU(n)=M(n,\mathbb{C})$. Could you point out my mistake? Thank you for your comment. –  Y. Fan May 16 '13 at 14:47
Probably you realized already, but the defining equation (see my answer) implies $U(n)$ is not a linear subspace (it is in fact defined by quadratic real algebraic equations), and this seems to have been your mistake. The same equation makes it easy to calculate the tangent space at $1$. –  Stephen May 16 '13 at 15:06

The tangent space to $U(n)$ at $1$ is the space of skew Hermitian matrices, not all matrices: this follows by differentiating the defining equation
$$A \overline{A}^t=1.$$ Now $\frac{c}{n} I_n$ is not skew Hermitian unless $c$ is imaginary. In fact, the definition of $U(n)$ implies $|\mathrm{det}(A)|=1$ for all $A \in U(n)$ (use that the determinant is multiplicative). So the determinant is a surjection onto the circle.
Thank you. I mistakenly think $U(n)$ is an open submanifold. In this case, its tangent space is the space of all matrices. However, $U(n)$ is a properly embedded submanifold. Therefore, it is closed. Thanks!! –  Y. Fan May 16 '13 at 15:05