bio | website | math.cornell.edu/m/People/… |
---|---|---|
location | Cornell University | |
age | 23 | |
visits | member for | 1 year |
seen | Apr 16 at 20:45 | |
stats | profile views | 263 |
Grad student at Cornell.
Apr 2 |
awarded | Yearling |
Mar 8 |
accepted | Why is “naive” definition of non-commutative spectrum bad? |
Mar 8 |
awarded | Nice Question |
Mar 5 |
asked | Why is “naive” definition of non-commutative spectrum bad? |
Jan 11 |
comment |
Understanding iterated integrals
Dear Matthew, thank you very much! The link does work, it's just the pdf file is huge (more than 60mb). |
Jan 11 |
comment |
Understanding iterated integrals
What is the name of the Deligne's paper you've mentioned? |
Dec 20 |
comment |
Examples of major theorems with very hard proofs that have NOT dramatically improved over time
Is this proof much simpler? If yes, do you know any English reference? I don't read French. Thank you! |
Dec 20 |
answered | Examples of major theorems with very hard proofs that have NOT dramatically improved over time |
Dec 17 |
comment |
Computer Algebra Errors
I think they have fixed it know. I tried and it says "True". |
Dec 5 |
awarded | Suffrage |
Dec 1 |
accepted | Morita theorem for simplicial rings |
Nov 24 |
comment |
Morita theorem for simplicial rings
Dear Martin, thank you very much for your answer! Can you, please, explain to me, why the Eilenberg-Watts theorem holds in the latter case? I have found a paper by Hovey "The Eilenberg-Watts theorem in homotopical algebra", and there he states the theorem for any cosmos $M$, but only for "nice" functors. For example, there should be a natural map $K\otimes FX\to F(K\otimes X)$ for any $K\in M$, $X\in Mod(A)$. Can you, please, explain, why is it the case when $M=simpAb$ and $C=Mod(B)$? |
Nov 20 |
revised |
Morita theorem for simplicial rings
deleted 14 characters in body |
Nov 19 |
asked | Morita theorem for simplicial rings |
Nov 1 |
accepted | Two definitions of modules in monoidal category |
Oct 31 |
comment |
Two definitions of modules in monoidal category
Yes, I know that. But this is far from being obvious (to me) that the axioms actually match up. Maybe I am just being stupid, and there is easy to see. I tried to write it down, but I didn't get anywhere. So I thought maybe there is some more high-powered way to say that everything works. |
Oct 31 |
revised |
Two definitions of modules in monoidal category
edited title |
Oct 31 |
asked | Two definitions of modules in monoidal category |
Oct 24 |
accepted | Chevalley restriction theorem for exterior algebras |
Oct 20 |
comment |
Chevalley restriction theorem for exterior algebras
Can you, please, explain to me, how to get the formula for $\bigwedge(\mathfrak{g}^*)$? I thought you should take $M=\mathfrak{g}$, $G$ acts on $M$ by conjugation. Then the tangent space to the orbit through $x\in M$ at $x$ will be $[\mathfrak{g},x]\subset \mathfrak{g}$, right? But then horizontal form $\omega$ is a form that annihilates all such tangent spaces. This means that $\omega$ annihilates $[\mathfrak{g},\mathfrak{g}]$. But if $\mathfrak{g}$ is semisiple, then $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so $\omega=0$. Can you, please, explain what I am doing wrong? |