One computer and one system-upgrade ago, I tried to convince people to set up their own MOPP (My Online Publications Page).
The essence of such pages is that they need an update, once in a while… I’m a bit embarrassed to admit it, but, I failed to re-install the package, following my own instructions.
Fortunately, there are plenty of good alternatives around, today. After playing a bit with bibtex2html, Ive settled for the bib2html perl script by Patrick Riley.
Included are well documented directions. The most important being that you have to do a
sudo ln -s /usr/bin/perl /usr/local/bin/perl51
and change the user-specific entries in the bib2html.conf configuration file (if you have already key.pdf files around, you can just drop them in the output-directory and set . as the paperfiledirlist, and they will be found and linked to automatically. If not, you can always include links in the bib-file). Further, if you want to link to web-pages of your coauthors, remember that the script expects you to use : LastName, FirstName | http://webpage even if you used a different convention in your bib-file.
The result can be seen here. For ebook-download-junks : Ive included again full PDF-files of all versions of my book (starting with version 1 from 1999 to the final published version of last year). They can best be found using the sort-by-type page.
- thank you Rupert, via the comments [↩]
-algebra A (not necessarily commutative). We will assign to it a strange object called the tangent-quiver 
, 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
of 
and between any two such vertices, say 
, the number of directed arrows from 
![\xymatrix{0 \ar[r] & S_w \ar[r] & V \ar[r] & S_v \ar[r] & 0}](/latexrender/pictures/79295bdb9289c15e0389919ccdcfde3e.gif "\xymatrix{0 \ar[r] & S_w \ar[r] & V \ar[r] & S_v \ar[r] & 0}")
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}](/latexrender/pictures/db75113e87fc6b4cc9024e320164e5b7.gif "\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}")
be a (commutative) affine variety with coordinate ring ![A = \C[X]](/latexrender/pictures/9ccc8dd12c038368ab65e6d2115c805a.gif "A = \C[X]")
then what is ![\C[X]](/latexrender/pictures/6ab887e2d620b4d8a3f151563dda413c.gif "\C[X]")
is commutative, all its finite dimensional simple representations are one-dimensional and there is one such for every point 
. Therefore, the vertices of 
corresponding to a point 
is just evaluating polynomials in 
then there are no non-split extensions between 
(a commutative semi-local algebra splits as a direct sum of locals), therefore in ![dim_{\C}~Ext^1_{\C[X]}(S_x,S_x) = dim_{\C}~T_x~X](/latexrender/pictures/1ee42834fa98fe925db931cc0ecf526f.gif "dim_{\C}~Ext^1_{\C[X]}(S_x,S_x) = dim_{\C}~T_x~X")
![A=\C[X]](/latexrender/pictures/720d51544089f662c6e2b42068360978.gif "A=\C[X]")
is the coordinate ring of an affine variety 
. So, in this case, the quiver 
of a finite group 
, 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 
of a finite quiver 
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{}}](/latexrender/pictures/e396315d4f3cf8c5deb4147af9b7699b.gif "\xymatrix{\vtx{} \ar[r] & \vtx{} \ar[r] & \vtx{}}")
matrices
where 
is the vertex idempotent. No doubt, you can guess what the tangent quiver 
will be, can’t you?