Algebraic Groups

So now we begin one of the great examples in algebraic geometry: algebraic groups. These are exceptionally nice, and we’ll talk about a couple of more general concepts before applying them to this case.

Before we can have any hope of talking about algebraic groups, we need to talk about products of varieties. We’re going to talk about our varieties without any reference to the affine or projective space that it is embedded in, and so we actually need to look at the Zariski topology on each of them.

Now, we can’t just say that the closed sets are the products of closed sets. Why not? Well, look at $\mathbb{C}^1\times\mathbb{C}^1$. The products of closed sets are just finite collections of points and vertical and horizontal lines. In particular, the parabola isn’t a closed set! This is bad, because we would LIKE to have $\mathbb{C}^1\times\mathbb{C}^1\cong\mathbb{C}^2$. The point set is right, we just need a good criteria for determining the topology.

So we’ll define the product of two varieties $V\times W$ to be the variety (and pair of morphisms $\pi_V:V\times W\to V,\pi_W:V\times W\to W$ such that given any other variety $Z$ with morphisms $f:Z\to V$ and $g:Z\to W$, there is a unique morphism $h:Z\to V\times W$ such that $f=\pi_V\circ h$ and $g=\pi_W\circ h$.

So what is this really saying? Well, we call the maps $\pi_V,\pi_W$ the projection maps, because the product has point set $\{(x,y)|x\in V,y\in W\}=V\times W$, and $\pi_V(x,y)=x$ and $\pi_W(x,y)=y$. So the universal property, that is, the map $Z\to V\times W$ is really to make sure that the topology works out.

We should note that though now we have $\mathbb{C}^n\times\mathbb{C}^m\cong\mathbb{C}^{n+m}$, we DON’T have $\mathbb{P}^n\times\mathbb{P}^m\cong \mathbb{P}^{n+m}$. This is VERY false. Though we do get that the product of affine varieties is affine and the product of projective varieties is projective. (This last bit requires a trick called the Segre Embedding which we won’t talk about).

So now on to algebraic groups. An algebraic group is a variety $G$ (affine, projective, quasi-projective…whatever) along with morphisms $m:G\times G\to G$, $\iota:G\to G$ and $1:\{pt\}\to G$. (the last one is just choosing a point of $G$, we’ll denote this point by $e$) These morphisms must satisfy a bunch of relations, specifically $m(x,m(y,z))=m(m(x,y),z)$, $m(e,x)=m(x,e)=x$ and $m(\iota(x),x)=m(x,\iota(x))=e$. These are just the axioms of a group, and so an algebraic group is just a variety which is also a group via morphisms. (This definition can be generalized to a group object in a category, and we will need this generality later, but for now, we’ll focus on varieties.)

So now, let’s mention a couple of examples. First, any group of matrices is an algebraic group. We can just look at $GL_n(\mathbb{C})$ to check this. Now, an invertible matrix is determined by the inequality $\det A\neq 0$, and so $GL_n(\mathbb{C})$ is by definition a quasi-affine variety in $\mathbb{C}^{n^2}$. However, we can do better. If we stick on another variable, so we’re working on $\mathbb{C}[x_{11},\ldots,x_{nn},y]$ as the coordinate ring of our affine space, we can then look at solutions to $y\det=1$, where $\det$ is the determinant polynomial in terms of the $x_{ij}$. This condition requires that the determinant be nonzero, and has no other requirements! Thus, $GL_n(\mathbb{C})$ is an affine variety in $\mathbb{C}^{n^2+1}$.

Now we need to check that inversion and multiplication are morphisms. From last time, we know that morphisms of affine varieties are the same as ring homomorphisms of the coordinate rings. So now if we take two matrices, the multiplication can be given by a bunch of polynomials in the coordinates, that is, $z_{ij}=\sum_k a_{ik}b_{kj}$, and so this is a morphism. Inversion is similar, though we need to divide by the determinant, but that’s ok, because it’s nonzero on our varieties. Thus, matrix groups are algebraic groups.

Anyone who knows a bit of the theory of Lie groups will recognize these as the standard examples. We’ll make this a bit more careful next, and that will be it for the day.

We say that a variety $X$ is nonsingular at a point $x\in X$ if $\dim T_x X=\dim X$. So that’s just that the tangent space and the variety are the same dimension. We say that $X$ is nonsingular if it is nonsingular at every point $x\in X$. If you’ll grant me, just for a moment, the implicit function theorem from analysis, then we can prove (though we won’t, because the proof isn’t terribly illuminating) that any nonsingular variety $X$ is, with the induced topology from the classical topology on $\mathbb{C}^n$ or $\mathbb{P}^n$, a complex manifold.

Now knowing this, to show that any algebraic group is a Lie group, we must only show that it is nonsingular at every point. First we’ll look at automorphisms of varieties. An automorphism is just an isomorphism $X\to X$. Now, we can compose and invert these perfectly well, and so the isomorphisms of $X$ form a group, which we will call $\mathrm{Aut}(X)$. We say that an automorphism group is transitive if for any $x,y\in X$, there exists $f\in \mathrm{Aut}(X)$ such that $f(x)=y$. That is, the automorphisms can move any point to any other.

Now, if there is an automorphism with $f(x)=y$, then we must have that $T_xX\cong T_yX$ by the map $df$, because we can invert $f$ do get $df^{-1}$, which gives that the map of tangent spaces is an isomorphism. Thus, any automorphism takes the nonsingular points to nonsingular points. So in particular, if $\mathrm{Aut}(X)$ is transitive, then if ANY point is nonsingular, all of them must be!

Later, when we talk about singularities a bit, we’ll prove that any variety has a nonempty open subset which is nonsingular, and so, in particular, has a nonsingular point. So this says that if the automorphism group is transitive, then the variety is nonsingular.

And now the final piece of the puzzle: the automorphism groups of algebraic groups are always transitive. To see this, we note that if we fix one of the points and look at $m(-,x):G\to G$ given by $y\mapsto yx$, this must be an isomorphism, and so is an isomorphism. So let $x,y\in G$. Then $m(-,x^{-1}y):G\to G$ takes $z\mapsto zx^{-1}y$, and so $m(x,x^{-1}y)=y$, so the automorphism group is transitive. So we now have that algebraic groups are nonsingular, and therefore manifolds. And so algebraic groups are Lie groups.

That’s all for tonight, I’m off until January, so there won’t be any new posts until then. Anyway, enjoy your holidays and a happy New Year to everyone.

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About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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8 Responses to Algebraic Groups

1. Anonymous says:

You define VxW using a universal property, which is of course the way it should be. BUT how do you proof existence in YOUR category of algebraic varieties? If V = {f = 0} and W = {g = 0}, then VxW is the vanishing locus of which polynomial?

2. Anonymous says:

No need to answer. It is of course {f = 0, g = 0} which is perfectly allowed in your category. I was confused.

3. Ravi says:

I was reading the page on Algebraic Groups and there are some missing equations (images) for example in the 6th and the 8th paragraphs. I have also noticed this in some of your other pages. It would be good to fix these – the pages are very informative – good for beginners like me…

4. Ravi, all of the equations are there as far as I can see. You might need to refresh the page, and possibly clear your cache first, because I know that once in a while when I look at math stuff on wordpress blogs, I get errors that are cleared up by that.

5. Ravi says:

Oops, all the equations _are_ there! Need to wait for all the images (math characters) to load…