Reputation
818
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
1 7
Newest
 Revival
Impact
~10k people reached

Jul
22
awarded  Revival
Jul
6
comment “Basis extension theorem” for local smooth vector fields
@AndrewD.Hwang Absolutely, in general $U$ may not be contractible and of course the proof would fall apart. I do mean $U\cap V$ for some sufficiently small, contractible open set $V$ containing $p$.
Jun
8
comment What are $E_\infty$-rings?
@AaronMezel-Gee Thanks. I'll check it out.
Jun
8
comment What are $E_\infty$-rings?
@AaronMazel-Gee I'm fairly well acquainted with the operad story. I was a bit more interested in the derived algebraic geometry part.
Jun
7
comment What are $E_\infty$-rings?
@AaronMazel-Gee This is a great summary of a lot of material! Do you have expository notes that expand on this? I know the pieces of the story, but it would be worth while to sit down and "connect all the dots" (as you clearly have).
Feb
24
revised Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem
deleted 9 characters in body
Feb
24
answered Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem
Feb
24
comment Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem
Also, the statement that $F$ has constant rank (as a smooth map between manifolds) means that the derivative $dF$ has constant rank as a map between tangent bundles. Hence, to fit it in to your context, $dF$ would need to have constant rank as a map of the tangent bundles over the TOTAL spaces $E$ and $E^{\prime}$.
Feb
24
comment Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem
I don't think you'll be able to use this. Since the total spaces of each bundle is a smooth manifold, certainly $F^{-1}(s(p))$ is an embedded submanifold ($s$ is the zero section, $p$ a point on the base). In fact, it's a subspace of the fiber. The problem is stitching these sub manifolds together smoothly.
Feb
24
comment Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem
What do you mean by the kernel of a smooth map. If you mean it's derivative, then this would be a special case of the claim where $E$ is the tangent bundle.
Feb
9
revised Question on tangent spaces
misspelled question
Feb
9
suggested approved edit on Question on tangent spaces
Feb
8
answered What to answer when people ask what I do in mathematics
Jan
30
comment Algebraic multiplicity of an eigen value
Algebraic multiplicity is just the degree of the linear term corresponding to $\lambda$ in the characteristic polynomial. Every matrix is similar to an upper triangular matrix and the determinant is invariant under similarity. Is this not enough to prove the claim?
Jan
12
comment Homomorphism between homotopy groups of spheres induced by the fibration and the multiplication map of $SO(n)$
@archipelago: Did you read my answer? I gave a large class of examples where $\lambda=0$.
Jan
12
answered Homomorphism between homotopy groups of spheres induced by the fibration and the multiplication map of $SO(n)$
Jan
11
comment Conditions for the integral to equal zero
The integral is nonzero even when $f(x)=1$. If your asking if $f(x)$ has to be bounded when the integral is finite, this is also not true. Consider, $f(x)=x$.
Jan
11
comment Nonlinear eigenvalue problem $Ax = f(c) x$
I'm a bit confused about the set up. If the $v_{i}'s$ are eigenectors with eigenvalues $b_{i}$, are they not precisely the eigenvectors for $f(c_{i})=b_{i}$. Is the base field real or complex? Is $A$ nice in any way? (e.g. diagonalizable, hermetian, unitary, orthogonal, real self-adjoint)
Jan
11
comment $k$-jets , submanifolds
I think your observation is the key step. $J^{1}(n,1)$ should just be the dual of $\mathbb{R}^{n}$, $\Sigma^{n-1}$ is all nonzero functionals and $\Sigma^{n}$ contains just the $0$ functional. Are these submanifolds? For the second part, perhaps you could use this characterization of $\Sigma^{n}$ along with the definition of a Morse function.
Jan
11
comment What does $d\zeta_1\wedge\cdots\wedge d\zeta_n$ mean in the context of Cauchy formula (on polydiscs)?
@TedShifrin: You are absolutely right. I, admittedly, did not read the problem carefully and assumed (a dangerous practice for mathematicians) that the resulting manifold would be complex.