PolyKnowMeAll
Reputation
365
Next privilege 500 Rep.
Access review queues
1 13
Impact
~3k people reached

45 Actions

 Sep24 awarded Autobiographer Aug21 awarded Popular Question Jul2 awarded Curious Jun4 awarded Nice Question Apr5 awarded Nice Question Feb18 comment Find an isomorphism between two graphs Yeah, seriously, dude, it's a cube. Just look at it. Label the vertices and you're done. Feb14 asked Variational characterization of gradient? Nov19 awarded Critic Nov15 accepted Why does it appear that Willmore energy is always zero? Nov15 awarded Commentator Nov15 comment Add vector on vector's tip and rotation Ok, then compute the length $a=|u+v|$; the vector you want is then $av/|v|$. Nov15 awarded Yearling Nov15 answered Cardioid: converting parametric form into polar coordinates Nov15 answered Add vector on vector's tip and rotation Nov15 comment Why does it appear that Willmore energy is always zero? This was a good exercise -- thanks! Nov15 answered Why does it appear that Willmore energy is always zero? Nov9 comment Why does it appear that Willmore energy is always zero? Hey Will - no book, but here's how I derived it. Stokes' says $\int_M d\alpha = \int_{\partial M} \alpha$ for $\alpha$ an $(n-1)$-form on an $n$-manifold. Then $\int_M df \wedge \star \alpha = \int_M d(f\star\alpha)-fd\star\alpha=\int_{\partial M} f\star\alpha - \int_M fd\star\alpha$ and the boundary integral in the final expression vanishes because there is no boundary. Letting $f=H$ and $\alpha=N^\flat$ we get the statement made in the original post. (Recall that $\nabla \cdot N = \star d\star N^\flat$ and $\nabla H = (dH)^\sharp$.) Nov9 accepted Fitting a quadratic polynomial of a special form Nov9 asked Why does it appear that Willmore energy is always zero? Nov20 comment Fitting a quadratic polynomial of a special form Ok, but still there are always a small number of solutions. It's surprising to me that these solutions are so hard to characterize.