Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In order to convince myself that the set $U(n)$ of unitary matrices (matrices with columns that are orthonormal under the complex inner product) is not an affine variety in $\mathbb{C}^{n^2}$, I need that the conjugate of an entry $\overline{x}_{ij}$ cannot be expressed using polynomials. If we say that complex polynomials have to be finite sums of monomials of the form $cx_1^{e_1}\dots x_n^{e_n}$ where $c \in \mathbb{C}$, $e_i \geq 0$, then is this really enough? It seems a little slippery.

share|improve this question
add comment

3 Answers

up vote 7 down vote accepted

That conjugation is not expressible in terms of polynomials comes from the Cauchy–Riemann equations, but that is not what you need to prove. You need to prove that the set $U(n)$ cannot be defined by a system of polynomial equations in the entries of the matrix.

Here is a simple proof. Consider the set $X$ of matrices $(a_{ij})$ such that $a_{ij}=\delta_{ij}$, except for $i=1, j=1$. Then $X$ is an affine variety in $\mathbb{C}^{n\times n}$. If $U(n)$ is an affine variety, then so is $Y=X \cap U(n)$. But $Y$ corresponds to the unit circle in $\mathbb C$. However, the unit circle is not an affine variety in $\mathbb C$, because those are either $\emptyset$, $\mathbb C$, or a finite set of points.

share|improve this answer
3  
Right. That is, the standard definition of $\text{U}(n)$ involves functions which are not polynomials, but this doesn't prove that some more esoteric definition can't be used which does. –  Qiaochu Yuan May 31 '11 at 13:43
    
Yes, that is a good point. Good, so the fact that the standard definition does not work by the Cauchy-Riemann equations answers my original question.. so how does one go about showing that is not possible to define $U(n)$ by polynomials, even with an alternative definition? –  Sputnik May 31 '11 at 13:56
3  
@Fahad, try embedding $U(n-1)$ into $U(n)$. In other words, find an algebraic subset of $U(n)$ that is isomorphic to $U(n-1)$. If $U(n)$ was an algebraic set, then $U(n-1)$ would be too, and you could go all the way down to $n=1$, which gets you back to $z \bar z=1$ not being an algebraic set over $\mathbb C$. –  lhf May 31 '11 at 14:03
    
Thank you, I will try that, though it sounds elaborate for the exercise I'm attempting. It is only the second exercise in An Invitation to Algebraic Geometry by Smith & Kahanpää so it seemed like they were after the conjugation idea, judging from the difficulty of other exercises and the level of content in that chapter. –  Sputnik May 31 '11 at 16:24
    
@Fadah, I've added a simple proof to my answer. –  lhf May 31 '11 at 18:23
add comment

Probably Perhaps the most conceptually straightforward explanation of why $U(n)$ is not an affine variety over $\mathbb{C}$ is the following: with respect to the complex topology (the usual topology -- not the Zariski topology!) $U(n)$ is compact, whereas it can be shown that a variety $V_{/\mathbb{C}}$ is such that $V(\mathbb{C})$ is compact (with respect to complex topology) iff $V$ is complete. In particular, a complete affine variety is finite, which $U(n)$ evidently is not.

Of course all of these facts require (nontrivial) proof. If I am remembering correctly, they are well treated in Shafarevich's text on algebraic geometry, for instance.

share|improve this answer
    
You're right, it is conceptually straightforward, though I can't say I have met any of those facts yet! I will keep the proof in mind for when I do. –  Sputnik May 31 '11 at 16:27
add comment

Here is a proof that $U(n)$ is not an affine variety in $\mathbb{C}^{n^2}$: If $X \subset \mathbb{C}^N$ is a smooth affine subvariety of $\mathbb{C}^N$, then, at every point of $X$, the tangent space to $X$ will be a complex sub-vector-space of $\mathbb{C}^N$. (Proof: The tangent space to a polynomial hypersurface is a complex sub-vector-space; the tangent space to an intersection is the intersection of tangent spaces. This argument also works for singular varieties if you use the Zariski tangent space.)

At the identity, the tangent plane to $U(n)$ is the Hermitian matrices. Mutiplyling a Hermitian matrix by $i$ does not give you a Hermitian matrix, so $U(n)$ is not an algebraic subvariety.

share|improve this answer
2  
To paraphrase this a bit, it seems to me that you are saying: if $U(n) \subset \mathbb{C}^n$ were a complex variety, then by homogeneity its set of $\mathbb{C}$-points would be a complex submanifold of $\mathbb{C}^n$ and thus its tangent spaces would be $\mathbb{C}$-subspaces...which they're not. –  Pete L. Clark May 31 '11 at 15:12
    
Thank you! This is a very nice proof but as with Pete's answer, I have to say I have not met that result yet, nor even looked at tangent spaces properly.. it is on the agenda though. –  Sputnik May 31 '11 at 16:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.