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Let $X$ and $Y$ be two subvarieties of $n\times n$ matrices. My question is that is there any condition to guarantee that there exits some matrix $g$ such that $Y=g^{-1} X g$? If such $g$ exists, then let's say $X$ is conjugate to $Y$. It is clear that the necessary condition is that $\dim X=\dim Y$ and $\deg X= \deg Y$.

The motivation of this problem is that if $X,Y$ are just two points, then we can use Jordan canonical form to check whether or not $X$ is conjugate to $Y$. But what if $X,Y$ are of higher dimension? For example, if $X,Y$ are two curves in the space of $n\times n$ matrices, when is $X$ conjugate to $Y$? Another simple case is that if $X,Y$ are linear subspaces of dimension 2, then when is $X$ conjugate to $Y$? In this case, one can rephrase the question as follows, consider the action of $GL_n$ on the Grassmannian $Gr(2,n^2)$ by conjugation. When are two points of $Gr(2,n^2)$ in the same orbit under this action?

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Where do you get the coefficients of your matrices? The answer will depend on the domain... –  Igor Rivin Mar 19 '14 at 18:57
Can you please give a specific example that you would like checked? Phrased in this way, the question is open to wide interpretation. –  Jason Starr Mar 21 '14 at 14:42

2 Answers 2

I may be completely off-base, and would be happy to be proved wrong, but I believe you are veering close to problems that are reputedly intractable. Namely, to simplify your simplest example even further, suppose $X$ consists of two commuting matrices $a, b$. Then you're asking about the classification of such pairs up to the action $g(a,b)=(gag^{-1}, gbg^{-1})$ of $\mathrm{GL}(n)$. Now, quoting Math Reviews:

In representation theory, a classification problem is called wild if it contains the problem of classifying pairs of matrices up to simultaneous similarity. It is known [see, for example, P. Gabriel; I. M. Gelʹfand and V. A. Ponomarev] that this problem contains that of classifying representations of an arbitrary quiver.

I believe it was Gel'fand who made the above the definition of a wild, i.e. "intractable" problem. His above-cited paper with Ponomarev starts:

There is extensive literature devoted to the problem of the canonical form of a pair of commuting linear transformations $a$ and $b$ in a finite-dimensional space relative to the transformations $a'=cac^{-1}$ and $b'=cbc^{-1}$, where $c$ is a non-singular linear transformation. We will show that attempts of direct determination of the canonical form do not make sense. Namely, the solution of this problem would imply the solution of the problem of classifying any finite number of arbitrary non-commuting linear transformations.

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Why conmuting? Classifying two matrices up to conjugacy is the prototypical wild problem. –  Benjamin Steinberg Mar 19 '14 at 22:46
@BenjaminSteinberg I've added a second quote, of Gel'fand-Ponomarev, which I believe answers your question. –  Francois Ziegler Mar 19 '14 at 23:04
Yes commuting operators is also wild. –  Benjamin Steinberg Mar 20 '14 at 1:15
I think the OP is asking not for a classification, but just for a computational criterion for when given $X$ and $Y$ are conjugate. It is "impossible" to classify similarity classes of pairs $(a,b)$, but given $(a,b)$ and $(a',b')$, it is fairly easy to see if they are similar: first check if $a'$ is similar to $a$, i.e., reduce to the case that $a'$ equals $a$, and then compare $gbg^{-1}$ to $b'$ for $g$ in the centralizer of $a$. –  Jason Starr Mar 21 '14 at 14:41

It's kind of a broad question. A good answer really depends on what you know about $X$ and $Y$. Suppose you are given the equations for both. Let $I_X$ and $I_Y$ be their respective ideals in the coordinate ring $R= k[x_{ij}]$ of the space $\mathbb{A}^{n^2}$ of $n\times n$ matrices. Then $G=GL_n(k)$ acts by $R$ compatible with the conjugation action on $\mathbb{A}^{n^2}=Spec R$, the ring $R^G$ is isomorphic to a polynomial ring in $n$-variables by classical invariant theory. We can view $Spec R^G\cong \mathbb{A}^{n}$ as the GIT quotient $\mathbb{A}^{n^2}//G$. A necessary condition for $X$ and $Y$ to be conjugate is that their images in the quotient coincides. In terms of a practical test which is almost equivalent, test whether $I_X\cap R^G=I_Y\cap R^G$. If it passes this test, then compare fibres of the projections of $X$ and $Y$ over these sets. Again this can be made pretty explicit, but I'll let you figure this out if you are interested.

Further comments As Francois points out, the problem seems to be intractable in general. However, I think one can say something useful under appropriate conditions. Let $U\subset \mathbb{A}^{n^2}$ be the locus of semisimple matrices. This is Zariski open. The above GIT quotient can be identified with the usual orbit space $\mathbb{A}^n = U/G$. I think this can be extended to $m$-tuples as follows. The ring of invariants of the $m$-fold product $(R\otimes\ldots \otimes R)^G$ is finitely generated (Processi) and I am pretty sure that its spectrum can be identified with the orbit space $(U^m)^G$. What this means concretely, is that conjugacy problem can solved, in principle*, for $m$ semisimple matrices. I can think this can be extended to cover irreducible varieties which meet $U$, but that involves more effort than I'm willing to put in right now. Perhaps this partly addresses Vanchinathan's question.

*This requires getting explicit generators of the ring of invariants...

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Suppose if further $X$ and $Y$ were embedded in a non-degenerate way in the affine space of $n\times n$ matrices, will the answer be yes? –  P Vanchinathan Mar 19 '14 at 23:41

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