Wilking's connectivity theorem says, $X$ is a positively curved manifold and $Y$ is a totally geodesic submanifold of codimension $k$,then $Y$ is $n-2k+1$ connected in $X$.Then follow the theorem can we get that $X-Y$ has homology up to $2k-2$?
1 Answer
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This is a consequence of Poincaré duality applied to the open $n$-manifold $X-Y$.
In particular, we have $H^i_c(X-Y)\cong H_{n-i}(X-Y)$ for all $i$, where $H^\ast_c$ denotes cohomology with compact supports. So it suffices to check that the compactly supported cohomology of $X-Y$ vanishes below degree $n-2k+2$.
There is a long exact sequence for compactly supported cohomology, which in this case looks like $$ \cdots \to H^i_c(X-Y)\to H^i_c(X)\to H^i_c(Y)\to H^{i+1}_c(X-Y)\to\cdots $$ and so the claim follows from the connectivity assumption on the inclusion map $Y\subseteq X$.
This all works with integral or mod 2 coefficients, depending on the orientability of the spaces involved.
Edit: The above argument works when $X$ and $Y$ are compact, which is the setting for Wilking's connectivity theorem mentioned by the OP (Acta Math. 191 (2003), no. 2, 259–297). In that case, compactly supported cohomology coincides with ordinary cohomology, and the map $H^i(X)\to H^i(Y)$ is an isomorphism for $i< n-2k+1$ and a monomorphism for $i=n-2k+1$ by naturality of the universal coefficient exact sequence and the five lemma.
It is false in general that a map of manifolds $f\colon\thinspace Y\to X$ which is $r$-connected induces an isomorphism $f^\ast\colon\thinspace H^i_c(X)\to H^i_c(Y)$ for $i< r$ and a monomorphism for $i=r$, because compactly supported cohomology is not homotopy invariant. Take $X=\mathbb{R}^n$ and $Y=\lbrace 0\rbrace$ for example.
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$\begingroup$ My algebraic topology is poor.Y is n−2k+1 connected in X,then homology H_k(Y) iso to H_k(X) when k<n-2k+1 and surjective when k=n-2k+1.So I think compact cohomology of X iso to Y when K>2k-1,surjective(?)when k=2k-1.Then by the long exact sequence,compact cohomology of X-Y =0 when k>=2k-2.So H_k(X-Y)=0 when k<=n-2k+2.Why I am wrong? $\endgroup$ Commented Sep 12, 2012 at 13:46
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$\begingroup$ It looks like I am claiming that since the inclusion map $f\colon Y\to X$ is $n-2k+1$ connected, the map $f^\ast\colon H^i_c(X)\to H^i_c(Y)$ is iso for $i<n-2k+1$ and mono for $i=n-2k+1$. This is certainly true if $X$ and $Y$ are compact, by the naturality of the universal coefficient sequence and the five lemma. I think it should be true more generally, but may require an argument involving direct limits. $\endgroup$ Commented Sep 12, 2012 at 14:27
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$\begingroup$ By the way, in your comment you seem to be composing the iso on homology with Poincaré duality to get an iso on cohomology with compact supports. But note that this introduces a degree shift, since $X$ and $Y$ have different dimensions. So you may have to use the UCT, or something similar. $\endgroup$ Commented Sep 12, 2012 at 15:25
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$\begingroup$ Oh,different dimensions!Thank you for your answer.I wonder whether we can put forward a result nearer to the manifold case. $\endgroup$ Commented Sep 13, 2012 at 8:18
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1$\begingroup$ My English is poor,Sorry that I have not made me clear.The conjecture for Alexandrov space is "has homology only up to 2k-2",which is far from the manifold case "n-2k+1 connected".I wonder whether we can get another conjecture for Alexandrov space case.But by now I have no idea. $\endgroup$ Commented Sep 14, 2012 at 6:01