14
votes
Accepted
What does Robert Stong mean when he says $H^*(MO(k))$ is a free Steenrod algebra in dimension less than $2k$?
First, let's clarify what the statement means:
The Steenrod algebra is a graded algebra, and $H^*(X)$ for any spectrum $X$ is a graded module over it. This means that the action $\mathcal{A}_2\otimes ...
- 10.8k
13
votes
Philosophy behind the Ricci flow
The Ricci flow behaves like a heat equation for the curvature with some additional reaction terms. Intuitively, the diffusion term acts to spread the curvature out and make the geometry more ...
- 6,001
6
votes
Accepted
Codimension zero embeddings and maps with small fibers
Isn't the following an example?
Build $M$ starting from the plane by attaching an infinite sequence of handles whose size decreases to zero.
Take for $N$ the disjoint union of $M_k$, where $M_k$ is ...
- 2,772
5
votes
Fundamental domain of an involution on a manifold
I will assume that you are working in the DIFF category, i.e. your manifold $X$ and involution $\tau$ of $X$ are smooth. It suffices t consider the case when $X$ is connected. Let $Y:=X/\tau$, $p: X\...
- 12.2k
5
votes
Cohomology of foliations and closed forms along the leaves
If $F$ is a regular foliation then we have a Lie algebroid on $M$ given as the (integrable) sub-bundle $L\subset T_M$ of vectors that are tangent to the leaves.
There is a Chevalley-Eilenberg complex $...
- 8,385
3
votes
Fundamental group of the complement of a codimension two submanifold
To your first question, the answer is yes.
Take a $k$-component trivial link in $S^n$, i.e. the boring, linear embedding
$$\sqcup_k S^{n-2} \to S^n$$
that is the boundary of a linear embedding
$$\...
- 44.3k
2
votes
Accepted
Smooth (locally trivial) fibration and submanifold
Let f:R^2 → R be the projection on the first factor and let L be the closed submanifold given by {(x,y)| x>0,xy=1 } union {(x,y)| y=-1 } . I think this is a counterexample.
- 201
1
vote
Cohomology of foliations and closed forms along the leaves
I am hardly an expert on this topic, but here's a construction.
Let $\Omega^k:=\Omega^k(M)$, and let$$F\Omega^k=\{\omega\in\Omega^k:\omega(v_1,\dots,v_k)=0,\quad v_1,\dots,v_k\in T_x\Sigma,\ \Sigma\...
- 2,792
1
vote
Chern class of roots of the canonical bundle
In general you have $c_1(L^{\otimes n}) = n c_1(L)$, because $c_1$ is a group homomorphism $c_1: \operatorname{Pic}(X) \to H^2(X,\mathbb Z)$. So if $L^{\otimes 2} = K$, then $c_1(L) = \frac 1 2 c_1(K)$...
- 1,286
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