David Speyer
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 2d comment Are all Hamiltonian planar graphs are 4 colorable? Does this imply all planar graphs are colorable? Consider the bipyramid: Take two tetrahedra and glue them face to face. The result is a planar graph with five vertices, six triangular faces and, also a $3$-cycle which is not a face. Apr 25 awarded Nice Answer Apr 14 comment minimum number of bases of a matroid, that comes from a convex polytope Nice! And equality is always achievable by a convex polytope -- Take $n-k+p-1$ points in generic convex position in $\mathbb{P}^{p-1}$, and make the other $k-p+1$ vertices independent to span $\mathbb{P}^{k-1}$. Apr 14 comment minimum number of bases of a matroid, that comes from a convex polytope @MoritzFirsching The nonbases of nonrealizable matroids tend to have little overlap with each other, so there can't be that many of them. Matroids with few bases tend to involve lots of points living in much lower dimensional flats, which is pretty easy to realize. All of the examples mentioned in mathoverflow.net/questions/212411/… , for example, are easily realizable. (Of course, this is all intuition until we actually prove something.) Apr 14 comment minimum number of bases of a matroid, that comes from a convex polytope @FedorPetrov Right, but phrase it as "What is the minimal number of bases of a matroid of rank $d+1$ on $n$ points, where all circuits have size at least $4$?" (A circuit of size 1 would be the zero vector, so not a point of $\mathbb{P}^d$; a circuit of size 2 would be two parallel vectors, so the same point of $\mathbb{P}^d$ counted twice; a circuit of size 3 would give three colinear points in $\mathbb{P}^d$, so not the vertices of a polytope.) It is possible that the optimal answer to this question is not realizable as the vertices of a convex polytope, but I would bet otherwise. Apr 14 comment Conjugacy classes of $SL_2(Z)$ Not the square root of the trace, but rather $\sqrt{t^2-4}$, where $t$ is the trace. Apr 11 revised Is $\max_{\|x\|_p=\|y\|_p=1} |\langle x, Ay\rangle|$ equivalent to $\max_{\|x\|_p=|} |\langle x, Ax\rangle|$ for symmetric $A$ & $p\geq 2$? added 379 characters in body Apr 11 revised Is $\max_{\|x\|_p=\|y\|_p=1} |\langle x, Ay\rangle|$ equivalent to $\max_{\|x\|_p=|} |\langle x, Ax\rangle|$ for symmetric $A$ & $p\geq 2$? added 879 characters in body Apr 11 answered Is $\max_{\|x\|_p=\|y\|_p=1} |\langle x, Ay\rangle|$ equivalent to $\max_{\|x\|_p=|} |\langle x, Ax\rangle|$ for symmetric $A$ & $p\geq 2$? Apr 11 comment Wrongful conviction Bayesian argument in need of integral-solving talent @YemonChoi Perhaps you should post your $SU(2)$ integrals! Ordinarily, I would think that an individual polynomial of the type you describe shouldn't be bad to integrate exactly, though asymptotics of a family might be hard to obtain. Apr 10 comment Is it true that; $\frac{{\partial f}}{{\partial x}}$ and $\frac{{\partial f}}{{\partial y}}$ don't have a common factor? No. Just take $p=0$ or $q=0$ and you get a counterexample. Or, for a slightly more interesting one, $-(1-x^2-y^2)^2 = (x^2+y^2) 2^2 -(1+x^2+y^2)^2$. Apr 8 awarded Nice Answer Apr 6 awarded Nice Answer Apr 5 answered Representations of the $3\times 3$ Heisenberg group Apr 4 comment Does $\mathbb C\mathbb P^\infty$ have a group structure? (1) "wow, this is much more complicated than I though it was" (2) "all of these examples involve contracting wild, fractal objects. Surely, it isn't so hard if all I want to deal with is contracting polyhedral subcomplexes of a simplicial complex." But I couldn't find any theorems which said contracting nice things is straightforward. This literature is very foreign to me, so I'd appreciate any pointers. Apr 4 comment Does $\mathbb C\mathbb P^\infty$ have a group structure? Could you point me to sources regarding "It is well-known that the quotient of a manifold by a contractible equivalence relation is homeomorphic to the original manifold." I spent a good deal of last night reading maths.ed.ac.uk/~aar/papers/daverman.pdf looking for theorems about when a quotient of a manifold is homeomorphic to the original manifold. I was left with two competing feelings (more) Apr 2 awarded Nice Answer Apr 2 awarded Necromancer Apr 2 comment Does $\mathbb C\mathbb P^\infty$ have a group structure? @TomGoodwillie This is easily seen to match Segal's construction (as described by Basu): your summands are the jump discontinuities of Segal's step functions. The problem is seeing that this is homeomorphic to $\mathbb{CP}^{\infty}$. Apr 2 comment Does homology have a coproduct? Just double subscripts to save space on the page.