Skip to content →

neverendingbooks Posts

coalgebras and non-geometry 3

Last
time we saw that the _coalgebra of distributions_ of a
noncommutative manifold can be described as a coalgebra
Takeuchi-equivalent to the path coalgebra of a huge quiver. This
infinite quiver has as its vertices the isomorphism classes of finite
dimensional simple representations of the qurve A (the coordinate ring
of the noncommutative manifold) and there are as many directed arrows
between the vertices corresponding to the simples S and T as is the
dimension of $Ext^1_A(S,T) $.

The fact that this
coalgebra of distributions is equivalent to the path coalgebra of
_some_ quiver is in the Kontsevich-Soibelman
paper
though it would have been nice if they had given reference for
this fact to the paper Wedge Products and
Cotensor Coalgebras in Monoidal Categories
by Ardizzoni or to
previous work by P. Jara, D. Llena, L. Merino and D. Stefan,
“Hereditary and formally smooth coalgebras”, Algebr.
Represent. Theory 8 (2005), 363-374. In those papers it is shown that a
coalgebra with coseparable coradical is hereditary if and only if it
is formally smooth if and only if it is a cotensor coalgebra of some
bicomodule.

At first this looks just like the dual version of
the classical result that a finite dimensional hereditary algebra is
Morita equivalent to the path algebra of a quiver (which is indeed what
the proof does) but again the condition that the coradical is
coseparable does not require the coradical to be finite dimensional…
In our case, the coradical is indeed coseparable being the direct sum
over all matrix coalgebras corresponding to the simple representations.
Hence, we can again recover the _points_ of our noncommutative manifold
from the direct summands of the coradical. Fortunately, one can
compute this huge coalgebra of distributions from a small quiver, the
_one quiver to rule them all_, but as I’ve been babbling about all of
this here [numerous
times](http://www.neverendingbooks.org/?s=one+quiver) I’ll let the
interested find out for themselves how you use it (a) to get at the
isoclasses of all simples (hint : morally they are the smooth points of
the quotient varieties of n-dimensional representations and enough tools
have been developed recently to spot some fake simples, that is smooth
proper semi-simple points) and (b) to compute the _ragball_, that is the
huge quiver with vertex set the simples and arows as described
above. Over the years I’ve calculated several one-quivers for a
variety of qurves (such as amalgamated free products of finite groups
and smooth curves). If you are in for a puzzle, try to determine it for
the qurve $~(\mathbb{C}[x] \ast C_2) \ast_{\mathbb{C}
C_2} \mathbb{C} PSL_2(\mathbb{Z}) \ast_{\mathbb{C} C_3}
(\mathbb{C}[x] \ast C_3) $ The answer is a mysterious
hexagon

Leave a Comment

doing the Perelman

Can I
suggest this addition to mathematical terminology?

doing a Perelman = making a voluntary retreat from the
math circuit to preserve one’s own well-being (either mental,
physical, scientific …).

As in : “I’m doing a
Perelman ever since that Oberwolfach meeting in 2002″

I
guess by now everyone has read the New Yorker-article by Sylvia Nasar
and David Gruber
Manifold Destiny. A legendary problem and the battle over who solved
it.
summarized in the accompanying drawing

In
case you never made it to the last page, here is the crucial paragraph

Perelman repeatedly said that he had retired
from the mathematics community and no longer considered himself a
professional mathematician. He mentioned a dispute that he had had years
earlier with a collaborator over how to credit the author of a
particular proof, and said that he was dismayed by the
discipline’s lax ethics.

“It is not people who
break ethical standards who are regarded as aliens,” he said.
“it is people like me who are isolated.”

We asked
him whether he had read Cao and Zhu’s paper. “It is not
clear to me what new contribution did they make,” he said.
“Apparently, Zhu did not quite understand the argument and
reworked it.”

As for Yau, Perelman said, “I
can’t say I’m outraged. Other people do worse. Of course,
there are many mathematicians who are more or less honest. But almost
all of them are conformists. They are more or less honest, but they
tolerate those who are not honest.”

3 Comments