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The question comes from Exercise 1.1.2 of the book "An Invitation to Algebraic Geometry". By definition the unitary group U(n) is the group of all complex matrix that satisfies $U^*U=I$. I know that since conjugate is involved the definition equations are not complex polynomials.

But this doesn't answer the question. To answer the question we have to prove that "there does not exist a set of polynomials whose common zero set is U(n)".
For odd n I have an idea: since the real dimension of U(n) is $n^2$, when n is odd it cannot be a complex variety whose real dimension should always be even(am I right here?). But what if n is even? Any hints will be appreciated.

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I think I've got the idea of @orangeskid answer. Let me edit the post and make it more detailed, so that others can understand.

First let's state 3 lemmas:

Lemma 1. Given an analytic function $f$ on $\mathbb{C}^n$, and suppose $W$ is a $n$ (real) dimensional linear subspace of $\mathbb{C}^n$ such that $\mathbb{C}^n\cong W+iW$, if $\left.f\right |_W\equiv 0$, then $f\equiv 0$.
Lemma 2. The Lie algebra $W=u(n)\subset \mathbb{C}^{n^2}$ of the Lie group $U(n)\subset GL(n,\mathbb{C})$ has a real dimension of $n^2$, and has the property that $\mathbb{C}^{n^2}\cong W+iW$.
Lemma 3. Given an analytic function $g: GL(n,\mathbb{C})\rightarrow \mathbb{C}$ that vanishes on $U(n)$, the exponential map exp transplants it to an analytic function $f=g\circ \exp: \mathbb{C}^{n^2}\rightarrow \mathbb{C}$ that vanishes on $u(n)$.

Once we have the 3 lemmas, it's easy: Suppose $g: \mathbb{C}^{n^2}\rightarrow \mathbb{C}$ is a polynomial vanishing on $U(n)$. Based on Lemma 3 we get an analytic function $f=g\circ \exp: \mathbb{C}^{n^2}\rightarrow \mathbb{C}$ vanishing on $u(n)$, so according to Lemma 1 and 2 we have $f\equiv 0$, which implies $g\equiv0$ on $\exp(\mathbb{C}^{n^2})=GL(n,\mathbb{C})$. Q.E.D.

So we need only to prove the lemmas.

Proof of Lemma 1:

Identifying $\mathbb{C}^n$ as $\mathbb{R}^{2n}$, choose an orthonormal basis $\{e_1, e_2, \dots, e_n \}$ of $W$. Then $\{e_1, \operatorname{J}(e_1), \dots, e_n, \operatorname{J}(e_n)\}$ is an orthonormal basis of $\mathbb{R}^{2n}$, where $\operatorname{J}$ is the almost complex structure of $\mathbb{C}^n$. Do a coordinate transformation so that these basis become the (real)coordinate basis of the coordinate $\{z_1, \dots, z_n\}=\{x^1+iy^1, \dots, x^n+iy^n\}$ and the conclusion reduces to the following one:

If the analytic function $f: \mathbb{C}^n \rightarrow \mathbb{C}$ vanishes on the real coordinates, that is, $f(x^1, \dots, x^n)\equiv 0$, then $f(z_1, \dots, z_n)\equiv 0$.
$\forall x^2, \dots, x^n$, let $g(z)=f(z, x^2, \dots, x^n)$, $g$ is analytic and $g(x)=f(x, x^2, \dots, x^n)\equiv 0, \forall x\in \mathbb{R}$. So $g\equiv 0$, that is, $f(z_1, x^2, \dots, x^n)\equiv 0, \forall z_1\in\mathbb{C}, x^i\in\mathbb{R}, i=2, \dots, n$. Continue the process we'll get $f(z_1, z_2, \dots, z_n)\equiv 0$. Q.E.D.

Proof of Lemma 2:
$u(n)$ is the group of all n × n skew-Hermitian matrices, having a real dimension of $n^2$. For any n × n complex matrix A, Write A as $$A=\frac{A-A^*}{2} + i\cdot(-i)\frac{A+A^*}{2}$$. Q.E.D.

Proof of Lemma 3:
This is trivial since exp is analytic. Q.E.D.

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  • 3
    $\begingroup$ In short, unitary groups are compact, but in the analytic topology, affine complex varieties are not. $\endgroup$ – AnonymousCoward Sep 16 '14 at 21:43
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    $\begingroup$ I can prove that a hypersurface V is not bounded: for almost every $z_0 \in \mathbb{C}^{n-1}$, $f(z_0,w)$ reduces to a polynomial of w. This polynomial has a root $w_0$, so we get a point $(z_0,w_0)\in V$. So the projection $(z,w)\mapsto z$ covers $\mathbb{C}^{n-1}$ with a finite set of exceptions. Since the image space is un-bounded, so does V. However how do we prove that any complex affine varieties are not bounded? $\endgroup$ – Xipan Xiao Sep 17 '14 at 14:52
  • $\begingroup$ @anonymouscoward $\endgroup$ – Xipan Xiao Sep 18 '14 at 13:58
  • $\begingroup$ @XipanXiao: affine complex varieties of dimension $n \ge 1$ are not compact in the analytic topology. One way to see this is by Noether normalization, which implies that there is a (finite) surjection to $\mathbb{C}^n$, but $\mathbb{C}^n$ is not compact $\endgroup$ – zcn Sep 21 '14 at 20:42
  • $\begingroup$ @zcn: Is there a way we don't need any further tools like Noether normalization? I'm asking because this exercise, is the second question in the book, just after the definition an affine variety is proposed. That makes me think this is quite an easy problem $\endgroup$ – Xipan Xiao Sep 21 '14 at 20:46
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Well, you can even prove that there does not exist a set of complex analytic functions on $M_n(\mathbb{C})$ whose zero set is $U(n)$. And not even locally. One shows the following: let $V$ an open subset of $GL(n,\mathbb{C})$ intersecting $U(n)$ and $f$ a complex analytic function on $U$ so that $f_{|V\cap U(n)} = 0$. Then $f =0$ on $V$. This is similar to $\mathbb{R}^m \subset \mathbb{C}^m$ and it can be reduced to this situation by using the map $\exp$ from $u(n)$ to $U(n)$.

Here is the main point: Assume that we have $P(z_1, \ldots, z_n)$ a polynomial in the complex variables $z_1$, $\ldots$, $z_n$ ( so, a complex polynomial function on $\mathbb{C}^n$). Assume that $P(x_1, \ldots, x_n) = 0$ whenever $(x_1, \ldots, x_n) \in \mathbb{R}^n$ ( that is $P_{|\mathbb{R}^n} = 0$). Then $P$ is the zero polynomial.

Note that it's not only that $P$ is zero on some subspace half the dimension but it's zero on a subspace $W$ so that $W \oplus i W = \mathbb{C}^n$. This does not happen for instance for the polynomial $P(z_1, z_2) = z_1$. Here the set of zeroes is not a real form of $\mathbb{C}^n$ but is in fact a complex subspace.

A more delicate related question is whether any $U(n)$ with $n$ even can have a complex structure (Not whether it's a complex Lie group, that's easily not possible).

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  • $\begingroup$ Taking $\mathbb{C}^2$ as an example, $f(z, w)=z$ is analytic. It's zero set is the plane $\{z=0\}$. However $f$ is not identically zero out the plane. So I don't understand your argument. Could you be more detailed? $\endgroup$ – Xipan Xiao Sep 18 '14 at 13:55
  • $\begingroup$ I think I got your idea and edited the original post for a detailed proof. Could you check whether I did that correctly or not? $\endgroup$ – Xipan Xiao Sep 21 '14 at 19:33
  • $\begingroup$ Hi, you are doing a great job. Here is one issue, about the $\exp$ map, it is only locally an analytic diffeo around $0\in u(n)$. This is why you have to prove a local result, which is not harder to prove: if $U$ open, connected, $U \cap \mathbb{R}^n\ne \emptyset$ and $f$ analytic zero on $U \cap \mathbb{R}^n$ then $f$ zero on $U$. So: "real forms" of complex things are dense (Zariski,Zariski-analytic). There is not much to change in the proof, I like it. $\endgroup$ – Orest Bucicovschi Sep 21 '14 at 22:28
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Inspired by all of you(@AnonymousCoward, @zcn, @orangeskid) I got a simple and elementary proof:
Suppose $U(n)$ is an affine variety defined by a set of polynomials $\{f_{\alpha}\}$. Fix any $n-1$ uni-modular complex numbers $\lambda_2, \dots, \lambda_n$ such that $|\lambda_i |=1, i=2,\dots, n$, denote $A(z)=\operatorname{diag}(z, \lambda_2,\dots,\lambda_n)$ a diagonal matrix. $\forall \alpha$, let $g_{\alpha}(z)=f_{\alpha}(A(z))$, then $g_{\alpha}(z)$ is a polynomial of $z$. And $g_{\alpha}(z)$ vanishes on the unit circle, since $A(z)$ is a unitary matrix if $|z|=1$. Hense $g_{\alpha}\equiv 0$.
That is, the whole complex line $\{A(z)\subset\mathbb{C}^{n^2} \left. \right |z\in\mathbb{C}\}$ is contained in the variety, which implies that it is not bounded. Q.E.D.

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  • $\begingroup$ You reduced it to $U(1)$. Very nice! $\endgroup$ – Orest Bucicovschi Sep 23 '14 at 16:48

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