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age 45
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Math professor at Cornell, PhD 1996.

I dabble in string theory, but haven't published anything physical in ages.


1d
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.
1d
awarded  Caucus
Sep
24
awarded  Autobiographer
Jul
31
comment Isomorphic Dual and Conjugate Representations of a Lie Algebra
Check it once and for all for the Lie algebra $End(V)$ itself. I'm voting to close as not research-level.
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
May
22
asked Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$?
May
22
awarded  Student
May
22
asked Partial sum of ${A \choose i} {B\choose n-i}$, when $B=-1$?