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visits member for 3 years, 9 months
seen May 17 at 16:21

Aug
24
comment Convergence in mean square from expected value/variance
Hmm, this is very simple. I think my mind was rebelling at the implication that this implies almost-sure convergence for a subsequence of ${X_n}$. The latter fact seems too good to be true...
May
22
comment Dividing by an almost-surely positive random variable
So Y/Z is not a random variable (a function), but it is integrable with respect to this measure? Is there a name for this kind of "function"?
Dec
6
comment Density of a function of random variables
Ahh!! I'm foolish. My integral for z < 0 should be $\displaystyle\int_{-z}^\infty$. Thanks Nate.
Dec
6
comment Density of a function of random variables
Nate, I had a typo, the $z>0$ and $z<0$ regions were flipped. I agree that my $z<0$ part is giving the problem, but I don't see how to fix it.
Dec
5
comment Ext & Complexes
In Vista 3.4.6, Weibel says "... the set of equivalence classes ... (if this is indeed a set)", and then claims that $Ext^1(A,B)$ is an abelian group. How can $Ext^1$ not be a set, and still be an abelian group?
Dec
2
comment Cohomology of $\mathcal O_X$ for toric varieties
Whoops, I fixed it.
Dec
2
comment Cohomological decomposition of tensor sheaves?
Oh wow, I thought $H^m(X, \mathcal O_X)=0$ for all $m>0$!!
Dec
2
comment Cohomological decomposition of tensor sheaves?
Why is it false for $\mathcal O_X$? The only non-vanishing cohomology is $H^0(X,\mathcal O_X)$, so doesn't the formula hold?
Nov
15
comment Quotient spaces and equivariant cohomology
I'm not very familiar with stacks, or with the details of GIT if $G$ is not a torus, but at least when it is the torus the GIT procedure removes the non-free points from $Y$ before quotienting. $X//G$ is supposed to give a scheme $Y$, and I assume that it is constructed thus to avoid having the quotient be a stack.
Nov
5
comment Why is stable equivalence necessary in topological K-theory?
I think there's something basic I'm missing. Stable equivalence is an equivalence relation on isomorphism classes of bundles. Say $E$ and $F$ are isomorphism classes. Then it appears to me that $E = F$ iff $E + G = F + G$ in K-theory. I suppose my question is, why is stable equivalence not "lies in the same isomorphism class"? In particular, what is an example of two non-isomorphic bundles such that when a trivial bundle is added to them, they become isomorphic?
Nov
4
comment Why is stable equivalence necessary in topological K-theory?
Hi Ryan, thanks for your clarification, but it's still murky for me. I don't see why $E \cong F \not \Leftrightarrow (E \oplus G) \cong (F \oplus G)$
Nov
3
comment Why is stable equivalence necessary in topological K-theory?
Are you saying that stable equivalence endows isomorphism classes with a formal inverse? That I don't see.
Nov
3
comment Why is stable equivalence necessary in topological K-theory?
Clarified, I hope.