# Samuel Tinguely

Geneva

Age: 26

I'm currently a PhD student in the university of Geneva. For the moment, I work mainly in algebraic topology.
 Mar 20 awarded Yearling Mar 20 awarded Yearling Jun 25 awarded Tumbleweed Jun 20 answered Quotient geometries known in popular culture, such as "flat torus = Asteroids video game" Jun 18 accepted Why is the equivariant Euler class a character ? Jun 18 comment Why is the equivariant Euler class a character ?Okay, fine, now I understand. Thank you for this neat answer ! =D Jun 18 awarded Yearling Jun 18 answered Why is the equivariant Euler class a character ? Jun 18 awarded Commentator Jun 18 comment Why is the equivariant Euler class a character ?I can only see the first sentence as a rephrasing of my question. Is the map you're talking about the identification between \$H^\ast_T(pt)\$ and \$ST^\ast\$ ? If yes, why must it map a weight to the corresponding line bundle over a point ? If not, what is it ? Jun 17 asked Why is the equivariant Euler class a character ? Jun 12 comment Why is the Lie derivative linear in the vector field?Yeah, exactly ! It's a great way to put it, thanks ! Jun 11 comment Why is the Lie derivative linear in the vector field?Oh, yes, I failed to see this. >_< Thank you ! =) Jun 11 accepted Why is the Lie derivative linear in the vector field? Jun 11 comment Why is the Lie derivative linear in the vector field?Well, yes, but how the to prove that this is linear ? And it is the Lie derivative with respect to the vector field generated by \$X\$ and the action of \$G\$. Jun 10 asked Why is the Lie derivative linear in the vector field? May 31 comment a counterexample of covering projectionJust to be sure, I suppose that by \$\vee X_n\$ you mean their disjoint union ? And what makes you think that `no open subset of \$X\$ is evenly covered by \$p\$' ? I think on the contrary that \$X\$ is evenly covered by an countable dicrete fibre. May 31 comment Polynomials are continuous with respect to the Zariski topologyI suppose you want to call the original polynomial \$F\$, and that you mean that the set of points such that \$F(x_1,...,x_n)-p=0\$ is the zero-locus of a polynomial, and hence closed ? (because it is certainly not finite) May 31 comment Polynomials are continuous with respect to the Zariski topologyThis looks more like a homework problem than a true question. Perhaps should it be re-tagged accordingly ? Also, what have you already tried ? May 29 answered Thom class: Why are the two definitions equivalent?