Allen Knutson
Reputation
231
Next privilege 250 Rep.
1 7
Impact
~1k people reached

• 0 posts edited
• 1 vote cast

# 23 Actions

 Feb 4 awarded Supporter Jun 21 awarded Commentator Jun 21 comment Smooth Parameterization of Set of Linear Subspaces That parametrization being, take $T:S\to S^\perp$ to its graph $\{(v,Tv)\ :\ v\in S\}$. (Note that Peter Michor's $S$ is not the original poster's $S$.) And by the way, "a canonical way of doing this" exactly means not "given a basis of the vector space". Jun 10 comment Hilbert Series of $\mathbb{C}^2$? Let $R=\mathbb C[x,y,z]$, and instead of the $R$-module $R/I$ consider its associated graded $R/\sqrt{I} \oplus \sqrt{I}/I$ (with the same Hilbert function; here $\sqrt{I} = (z)$). The first is your plane $R/(z)$. The second is $R$-isomorphic to $R/(y,z)$ under the map $1 \mapsto z$, $R/(y,z) \to \sqrt{I}/I$, and $R/(y,z)$ is the coordinate ring of this $x$-axis. Note that the map $1\mapsto z$ changes the grading by $1$; this accounts for the $t^1$ in the numerator. Jun 10 comment Hilbert Series of $\mathbb{C}^2$? Actually, it's an embedded line (as the $x$-independence makes obvious), along $y=z=0$. Note that the difference in Hilbert series is $t/(1-t)$, the $1/(1-t)$ factor from the fact that it's one-dimensional, and the $t$ because it's "next to" the $y=z=0$ line in the big component. When dealing with monomial ideals, it's instructive to look at the set of monomials not in the ideal; your second one has a quadrant $\{x^i y^j\}$ worth, but the first has also a ray $\{ z x^i \}$. Jan 10 awarded Yearling Dec 20 comment Does an arbitrary product of $f$ and $f^\dagger$ belong to a universal enveloping algebra of the Heisenberg algebra? For somebody learning the subject, I think saying there's "no essential difference" between, e.g., a finite-dimensional and infinite-dimensional vector space is very confusing. "The representation theory is the same" is good and one should leave it at that. Dec 20 awarded Caucus Sep 24 awarded Autobiographer Jun 11 comment Explanation for a line from a MathOverflow answer Well, the key step is equivariant Kunneth, which I'm happy to apply if e.g. $H^*_S$ and $H^*_H$ are concentrated in even degree, e.g. if $S,H$ are connected. Without that I become squeamish about equivariant Kunneth. Apr 13 comment Intuition? how the author reaches the answer? Say $\vec w=(w_1,\ldots,w_8)$. Then declaring $x_2 x_4$ to be the initial term of $x_1 x_5 - x_2 x_4$, with respect to the $\vec w$ order, gives the linear inequality $w_2 + w_4 > w_1 + w_5$. In your first question, you need to decide which of the two terms in each binomial you want to be the leading term, in such a way that the resulting S-polynomials reduce to 0. But then you need to be sure that there is a term order that actually gives those choices of initial term. So find a $\vec w$ satisfying the three attendant linear inequalities. Apr 11 answered Intuition? how the author reaches the answer? Mar 15 comment Explanation for a line from a MathOverflow answer I think it's a mistake to explicitly include $EG$ factors in one of these deeper calculations. Accept that $H_G(X) = H(X/G)$ for free actions and then black-box that away. Partly because, when you move beyond $H^*()$ to $K()$ you don't use the Borel construction any more. Mar 14 awarded Teacher Mar 14 answered Explanation for a line from a MathOverflow answer May 23 comment Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$? ${a\choose k} = a(a-1)\cdots (a-k+1)/k!$ for $k\geq 0$, so ${-n\choose k} = (-1)^k {k+n-1\choose k}$. May 23 comment Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$? Yes, and I did, but not enough to guess an answer. Examples added. May 23 revised Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$? added 3756 characters in body; added 25 characters in body May 22 awarded Editor May 22 revised Partial sum of ${A \choose i} {B\choose n-i}$, when $B=-1$? added 173 characters in body