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M-geometry (1)

By lieven

Take an affine \C-algebra A (not necessarily commutative). We will assign to it a strange object called the tangent-quiver \vec{t}~A, compute it in a few examples and later show how it connects with existing theory and how it can be used. This series of posts can be seen as the promised notes of my talks at the GAMAP-workshop but in reverse order… If some of the LaTeX-pictures are not in the desired spots, please size and resize your browser-window and they will find their intended positions.

A vertex v of \vec{t}~A corresponds to the isomorphism class of a finite dimensional simple A-representations S_v and between any two such vertices, say v and w, the number of directed arrows from v to w is given by the dimension of the Ext-space

dim_{\C}~Ext^1_A(S_v,S_w)

Recall that this Ext-space counts the equivalence classes of short exact sequences of A-representations

\xymatrix{0 \ar[r] & S_w \ar[r] & V \ar[r] & S_v \ar[r] & 0}

where two such sequences (say with middle terms V resp. W) are equivalent if there is an A-isomorphism V \rightarrow^{\phi} W making the diagram below commutative

\xymatrix{0 \ar[r] & S_w \ar[r] \ar[d]^{id_{S_w}} & V \ar[r] \ar[d]^{\phi} & S_v \ar[r] \ar[d]^{id_{S_v}} & 0 \\\ 0 \ar[r] & S_w \ar[r] & W \ar[r] & S_v \ar[r] & 0}

The Ext-space measures how many non-split extensions there are between the two simples and is always a finite dimensional vectorspace. So the tangent quiver \vec{t}~A has the property that in all vertices there are at most finitely many loops and between any two vertices there are a finite number of directed arrows, but in principle a vertex may be the origin of arrows connecting it to infinitely many other vertices.

Right, now let us at least motivate the terminology. Let X be a (commutative) affine variety with coordinate ring A = \C[X] then what is \vec{t}~A in this case? To begin, as \C[X] is commutative, all its finite dimensional simple representations are one-dimensional and there is one such for every point x \in X. Therefore, the vertices of \vec{t}~A correspond to the points of the affine variety X. The simple A-representation S_x corresponding to a point x is just evaluating polynomials in x. Moreover, if x \not= y then there are no non-split extensions between S_x and S_y (a commutative semi-local algebra splits as a direct sum of locals), therefore in \vec{t}~A there can only be loops and no genuine arrows between different vertices. Finally, the number of loops in the vertex corresponding to the point x can be computed using the fact that the self-extensions can be identified with the tangent space at x, that is

dim_{\C}~Ext^1_{\C[X]}(S_x,S_x) = dim_{\C}~T_x~X

That is, if A=\C[X] is the coordinate ring of an affine variety X, then the quiver \vec{t}~A is the set of points of X having in each point x as many loops as the dimension of the tangent space T_x~X. So, in this case, the quiver \vec{t}~A contains all information about tangent spaces to the variety and that’s why we call it the tangent quiver.

Let’s go into the noncommutative wilderness. A first, quite trivial, example is the group algebra A = \C G of a finite group G, then the simple A-representations are just the irreducible G-representations and as the group algebra is semi-simple every short exact sequence splits so all Ext-spaces are zero. That is, in this case the tangent quiver \vec{t}~A in just a finite set of vertices (as many as there are irreducible G-representations) and no arrows nor loops.

Now you may ask whether there are examples of tangent quivers having arrows apart from loops. So, take another easy finite dimensional example : the path algebra A = \C Q of a finite quiver Q without oriented cycles. Recall that the path algebra is the vectorspace having as basis all vertices and all oriented paths in the quiver Q (and as there are no cycles, this basis is finite) and multiplication is induced by concatenation of paths. Here an easy example. Suppose the quiver Q looks like

\xymatrix{\vtx{} \ar[r] & \vtx{} \ar[r] & \vtx{}}

then the path algebra is 6 dimensional as there are 3 vertices, 2 paths of length one (the arrows) and one path of length two (going from the leftmost to the rightmost vertex). The concatenation rule shows that the three vertices will give three idempotents in A and one easily verifies that the path algebra can be identified with upper-triangular 3 \times 3 matrices

\C Q \simeq \begin{bmatrix} \C & \C & \C \\\ 0 & \C & \C \\\ 0 & 0 & \C \end{bmatrix}

where the diagonal components correspond to the vertices, the first offdiagonal components to the two arrows and the corner component corresponds to the unique path of length two. Right, for a general finite quiver without oriented cycles is the quite easy to see that all finite dimensional simples are one-dimensional and correspond to the vertex-idempotents, that is every simple is of the form S_v = e_v \C Q e_v where e_v is the vertex idempotent. No doubt, you can guess what the tangent quiver \vec{t}~A = \vec{t}~\C Q will be, can’t you?

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5 comments
Posted in geometry
Written on Sat, 15 September 2007 at 5:47 pm
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5 Responses to “M-geometry (1)”

  1. Greg Says:
    September 15th, 2007 at 8:01 pm

    It seems like this is the ‘first-order’ approximation to the representation theory of A, in the way that the K-theory is the ‘zeroth-order’ approximation.

    What I mean is, K(A) throws aways all the representation theory info in the Exts, whereas the tangent quiver seems to throw away all the representation theory info in Ext^n, n > 1. Are there similar constructions that yield algebras that measure Ext^n, for n greater than some integer? And, if so, can they form a directed system with a defined limit, some sort of ‘pro-finite’ approximation of the original algebra?

  2. lieven Says:
    September 15th, 2007 at 8:21 pm

    I guess the object you are interested in is the A_{\infty} structure on the extension-algebra Ext^{\ast}_A(M,M) for all semisimple A-modules M. A good paper on this is by Bernhard Keller “Introduction to Ainfinity algebras and modules” available via the arXiv (i think) or via his homepage.

    Im mostly interested in algebras where the first 2 terms of the approximation scheme are enough, so called formally smooth algebras or quasi-free algebras, so in my cases the quiver is all I need but I know it can be extended…

    A final comment about TeX. You can use TeX in comments but instead of $latex as opening tag use [ tex ] (without the spaces) and the closing $ should be replaced by [ /tex ] (sorry about that). (Ive edited your comment to get the TeX properly displayed, in fact better than mine because of the gray background…)

  3. javier Says:
    September 17th, 2007 at 11:03 am

    This may be well known, but I’ll ask anyway, When you mention the equality dim_{\C}~Ext^1_{\C[X]}(S_x,S_x) = dim_{\C}~T_x~X between the dimension of the [text]Ext[/tex] space and the tangent space in the classical setting. Is this equality always true, or do you require smoothnes of the variety X at the point x. In other words, can singularities be characterized by some property of the Ext spaces? Some other questions came while (re)reading this, but I’ll save them for later.

  4. lieven Says:
    September 17th, 2007 at 11:21 am

    Javier, I think X being reduced is enough. In general if M is an n-dml A-representation and the variety rep(n) of all n-dml A-reps is reduced, then the Ext-space is the normal space in M to the GL(n)-orbit of M. Here, A is commutative, n=1 and so X=rep(1) and the GL(1)-action is trivial, so it is just the tangent space to X.

    Edited a bit later : of course you do not need the general result in this case. If x is the point corresponding to the algebra map

    \rho~:~\C[X] \rightarrow \C

    (evaluation in x) then the Ext-space is just the space of all \rho-derivations and so coincides with the tangent space.

  5. javier Says:
    September 21st, 2007 at 4:36 pm

    But, if the variety X has a singularity at the point x, what is exactly the ‘tangent space’? I mean, the ‘tangent’ thing to a singularity is not a vector space at all, is it?

    Btw, why the ‘M’ in ‘M-geometry’? Does it have any concrete meaning?

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