Given an algebraically closed field $\mathbb{F}$ of characteristic $0$ and a closed subgroup $G$ of $GL_n(\mathbb{F})$. Let $\{g_1,\ldots, g_r\}$ be a Gr"obner basis for the correpsonding ideal $\mathcal{I}(G)$. I want to find the irreducible component of $1$ (that means i want to find generators for $\mathcal{I}(G^\circ)$). Is there a better way than computing the primary decomposition, pick the component, which contains $1$, and compute its radical?

  • $\begingroup$ How is $G$ given? $\endgroup$ – Mariano Suárez-Álvarez Nov 2 '10 at 15:28
  • $\begingroup$ It will already be radical. In characteristic zero, all group schemes are reduced. $\endgroup$ – David E Speyer Nov 2 '10 at 15:29
  • $\begingroup$ It would be helpful to mention why you want to do this concrete search, as David also indicates in his answer. Computation with defining polynomials of a linear algebraic group is fairly uncommon. $\endgroup$ – Jim Humphreys Nov 2 '10 at 22:24

I've never seriously attempted a computation like this, so don't take this answer too seriously. I'm curious where this problem comes from for you; I don't often encounter matrix groups given by defining equations. The only example I can think of is where one specifies some homogeneous polynomials and asks for the subgroup of $GL_n$ preserving them.

That said, I would start out by computing the Lie algebra of $G$. This is easy to do from defining equations. The Lie algebra of $GL_n$ is spanned by $\partial/\partial x_{ij}$ for $1 \leq i,\ j \leq n$. If $f_1$, $f_2$, ...., $f_r$ are generators for the ideal of $G$ then the Lie algebra of $G$ is the set of those vectors $\sum c_{ij} \partial/\partial x_{ij}$ such that $\sum c_{ij} (\partial f_k/\partial x_{ij})|_{\mathrm{Id}}$ is zero for $k=1$, ..., $r$. (You don't need a Grobner basis for this; any generating set works.) So, if you have $r$ defining equations, you can get the Lie algebra by solving $r$ linear equations in $n^2$ variables.

Why would I do this? Well, if the Lie algebra is $0$-dimensional, you win! The connected component of the identity is a point. More generally, if you recognize the Lie algebra as some classical Lie algebra you are familiar with (for example, if it's abelian) that might help you find equations.

So, what if you get the Lie algebra and it is unenlightening? Here is an idea, though I haven't tried it out. My strategy is going to be to try to find a function $z$ which is constant on connected components of $G$ and has no other special properties.

What will I do with $z$? Let $z_1$ be the value of $z$ at the identity. So $z-z_1$ vanishes on the connected component of the identity. If $z$ is generic, then $z - z_1$ will not vanish on any other component. So the ideal of the connected component of the identity is generated by the ideal of $G$, plus $z-z_1$.

So, how do I find $z$? Here I am facing a tradeoff between presenting methods which are computationally efficient, and methods which are guaranteed to work. I'll go with the latter route, but, if you actually are doing large examples, come back and ask some more specific questions.

Chose a positive integer $N$. The vector space of polynomials on $GL_n$ of degree $\leq N$ is stable under the left and right actions of $GL_n$ on itself, and hence for the action of $\mathrm{Lie}(G)$. The condition that a polynomial $f$ be annihilated by the action of a vector $v$ in $\mathrm{Lie}(G)$ is linear. So, by solving lots of linear equations, we can all functions on $G$ which are annihilated by $\mathrm{Lie}(G)$ and are restrictions of polynomials of degree $\leq N$. Since you have a Grobner basis, writing down the individual equations is not bad, although I suspect that their size will grow quite rapidly. As soon as you find such a function which is not the constant function, try it out in the role of $z$.

Of course, there is the risk that you already are working with a connected group, in which case you can search forever without finding a nonconstant $z$. So it would be best to check for this before going too far. Anyone have a bright idea how to do this, other than the original suggestion of computing primary decomposition?

  • $\begingroup$ Took me a while to understand why this works (e.g. $z$ exists and has the desired properties). I have some questions left: 1) How do you know that you picked the right $z$ without knowing the connected components? 2) You do this calculation with a fixed $N$ and increase it step by step until you found $z$, right? If yes: Can you bound $N$ (maybe by the degree of $f_i$)? $\endgroup$ – yell Nov 8 '10 at 13:57
  • $\begingroup$ Good questions! In both cases, I don't know. I had hoped to find a clever way to compute the entire ring of locally constant functions but, without computing connected components, it is not clear to me when I've succeeded. How hard is computing primary decomposition, anyway? $\endgroup$ – David E Speyer Nov 8 '10 at 14:57
  • $\begingroup$ By the way, there are several more clever things that I could do if I knew something about the Lie algebra in question. If $\mathfrak{g}$ is abelian, then I would just take a random high degree polynomial and project onto its weight zero component, rather than working with the whole space of degree $N$ polynomials. If $\mathfak{g}$ is reductive then I could in principle do the same thing, using the Casimir operator en.wikipedia.org/wiki/Casimir_operator , although I don't know how doable this is in practice. $\endgroup$ – David E Speyer Nov 8 '10 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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