Reputation
6,377
Next privilege 10,000 Rep.
Access moderator tools
Badges
2 26 69
Impact
~139k people reached

1d
revised Injectivity/Surjectivity of $F_A :=\frac{d}{dt} +A(t) $ for a hyperbolic path of matrices $A(t)$ on $H^1 $
parsing and type-setting
2d
comment CW 4 manifolds with single 4 cell
Unfortunately, it doesn't help (or isn't known) to get a CW-complex when $M$ is a non-smoothable 4-manifold.
2d
comment CW 4 manifolds with single 4 cell
The paper does not quite suffice when $M$ is a non-smoothable 4-manifold. From my link, it is known if $M$ is furthermore simply-connected.
Apr
27
comment CW 4 manifolds with single 4 cell
This was already asked (by myself) in more generality: mathoverflow.net/questions/120799/… . In particular, it answers both your questions!
Apr
26
revised Index of Modified Dirac Operator
added 2 characters in body
Apr
25
answered Index of Modified Dirac Operator
Apr
25
comment How to understand Taubes' moduli space of holomorphic curves?
Sorry I meant to say the branched covers will have positive (formal Fredholm) index, which increases by the number of branch points. Already that disagrees with the actual dimension of the moduli (which would increase by twice the number of branch points, due to the 2-dimensional movement of the branch points along the surface). Moral: expect no transversality for branched covers.
Apr
24
comment How to understand Taubes' moduli space of holomorphic curves?
It's the index constraint, branched covers have negative index. In terms of (co)homology classes, if the $J$-holomorphic curve is Poincare dual to a multiple of a cohomology class $e\in H^2(X)$ with $e\cdot e=0$ (so $PD(e)$ is represented by a self-intersection zero torus) then the curve is an unbranched cover of that embedded torus. As you mention, the SW solutions mainly recognize the subvarieties (image of the curves).
Apr
23
revised How to understand Taubes' moduli space of holomorphic curves?
added 82 characters in body
Apr
23
answered How to understand Taubes' moduli space of holomorphic curves?
Apr
23
reviewed Approve Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
Apr
21
answered Group cohomology of the cyclic group
Apr
19
reviewed Reject What is this Lie algebra?
Apr
19
revised Almost Complex Structures: 'Tame' versus 'Compatible'
simple rephrasing + add tag
Apr
15
comment Are knots determined by their complements within a homotopy class?
I am very confused on how doubling works here, because 1) we change the complements, and 2) I don't see what stops the knots from moving through the double to become isotopic. Am I correct in the setup that we first take $X:=S^3-\text{nbhd}(J)$ where $J$ is one component of the Whitehead link $J\cup K$ and then consider $K$ (along with some homotopic knot $K'$) inside $M=D(X)$? If so, then granted $J\cup K$ is not isotopic to $J\cup K′$ in $S^3$, I don't see how to use this to prove that $K$ is not isotopic to $K′$ in $M$. That is, why $K\cong K'$ in $M\;\Rightarrow\;K\cong K'$ in $X$?
Apr
8
comment Index of Modified Dirac Operator
I'm assuming $M$ is compact, and I'm assuming $\nabla f$ makes sense as a map $S^+\to S^-$ (I don't immediately see why it is). The point is that $D$ is Fredholm, and the space of Fredholm operators is open in the (Banach) space of bounded linear operators. So $ind(D)=ind(D+K)$ for any compact operator $K$ (hence $s$ can be anything), and if $K$ is not compact then under mild assumptions $ind(D)=ind(D+sK)$ for $s$ sufficiently small.
Apr
5
awarded  Popular Question
Mar
23
reviewed Reject spectrum of the compression of a selfadjoint operator
Mar
8
comment Obstructions to symplectically embedding compact manifolds of dimension $4$ or higher
Obstructions for $(M^4,\sigma)\hookrightarrow(X^4,\omega)$ are highly studied, using the notion of "capacities".
Mar
5
awarded  Favorite Question