Talk:Tutte polynomial

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Isn't the bit about the Jones polynomial all wrong?

The current text about the hyperbola xy=1 is completely at odds with Propos 1 of [ref], which in a later example clearly states that V_{D(G)}( s ) = ( s^{-1/2} + s^{1/2} )^{|E|-|V|+1} s^{1-|E|-|V|} T_G( s^2, (s^2+1)/(s^2-s) ).

[ref] Francois Jaeger, Tutte Polynomials and Link polynomials, Proc. Am. Math. Soc. Vol 103, No 2, June 1988 —Preceding unsigned comment added by 82.33.119.244 (talk) 18:40, 20 February 2010 (UTC)[reply]

Typesetting

The typesetting in this article needs fixing. 86.145.57.232 (talk) 11:33, 24 January 2014 (UTC)[reply]

Looks ok to me. What, specifically, do you see wrong? —David Eppstein (talk) 17:07, 24 January 2014 (UTC)[reply]

Wrong formula

I think this formula is wrong: $Z(G)=2\left(e^{-\alpha }\right)^{|E|-r(E)}\left(4\sinh \alpha \right)^{r(E)}T_{G}\left(\coth \alpha ,e^{2\alpha }\right).$ Isn't this should be $Z(G)=\left(2e^{-\alpha }\right)^{|E|-r(E)}\cdots$ ? 133.11.30.75 (talk) 16:27, 24 August 2014 (UTC)[reply]

Well-defined?

Why even mention well-definedness in the definition? There is no choice of representatives being made nor anyting that would potentially make the definition ambigous. Instead, I suggets it should be mentioned that the $\sum$ has only finitely many summands (hence indeed produces a polynomial) because the graph is finite. Or maybe this was what was meant with well-definedness in this context?--Hagman (talk) 12:02, 31 August 2014 (UTC)[reply]

There are several ways of defining the Tutte polynomial, for example one might start from the contraction-deletion definition. For some of these definitions there are choices, for example, the order of edges to operate on, which make it non-obvious that the answer is well-defined, in the sense of being independent of the various choices. Starting from the Whitney rank polynomial definition makes it clear that it is well-defined, because there are no choices to make, hence the comment in the text. Of course it is now not obvious that it is a contraction-deletion invariant. Deltahedron (talk) 12:36, 31 August 2014 (UTC)[reply]