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Let $Y\subset X$ be a codimension $k$ proper inclusion of submanifolds. If we choose a coorientation of $Y$ inside of $X$ (that is, an orientation of the normal bundle), then we get a class $[Y]\in H^k(X)$. If $X$ and $Y$ are oriented, then $[Y]$ may be defined as the fundamental class of $Y$ in the Borel-Moore homology of $X$, which is isomorphic to the cohomology of $X$. What is the simplest definition of $[Y]$ in the general case (where $X$ and $Y$ are not necessarily oriented)?

Note that a simple generalization of this question would be to ask how to define the pushforward in cohomology along a proper oriented map. (Then $[Y]$ would simply be the pushforward of $1\in H^0(Y)$.) I would be happy to know the answer to this more general question, but I asked the simpler version to be as concrete as possible.

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4 Answers 4

up vote 7 down vote accepted

The easiest definition is via the Pontrjagin--Thom construction (I think). Let $N$ be a tubular neighbourhood of $Y$, isomorphic to the normal bundle. Let $X'$ be the space obtained from $X$ by collapsing the complement of $N$ to a point. Then $X'$ is isomorphic to the Thom space of the normal bundle, and if the normal bundle is oriented then there is the Thom isomorphism: $$ H^\bullet(Y,\mathbf Z) \cong H^{\bullet+\dim X - \dim Y}(X',\mathbf Z).$$ Composing with the natural $H^\bullet(X',\mathbf Z) \to H^\bullet(X, \mathbf Z)$ gives the map you want.

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Your construction for the oriented case extends to the unoriented case just by using twisted coefficients.

Every manifold $Y$ has a fundamental class $[Y]\in H_\ast^{\mathrm{lf}}(Y;\mathfrak o_Y)$ (where $\mathfrak o_Y$ is the orientation sheaf of $Y$). The coorientation of $Y\subseteq X$ gives an identification of orientation sheaves $\mathfrak o_Y\xrightarrow\sim\mathfrak o_X$ over $Y$, and thus there is a pushforward map: $$H_\ast^{\mathrm{lf}}(Y;\mathfrak o_Y)\to H_\ast^{\mathrm{lf}}(Y;\mathfrak o_X)\to H_\ast^{\mathrm{lf}}(X;\mathfrak o_X)$$ Now use the Poincare duality isomorphism $H^\ast(X;\mathbb Z)\xrightarrow{\cap[X]}H_\ast^{\mathrm{lf}}(X;\mathfrak o_X)$.

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Thanks, that's a nice perspective! –  Nicholas Proudfoot Dec 11 '13 at 16:39

This is an answer to your more general question about how to define $$f_\ast:H^\ast(X;\mathbb{Z})\to H^{\ast+\dim(Y)-\dim(X)}(Y;\mathbb{Z})$$ when $f: X\to Y$ is a proper oriented map.

For $\ell$ sufficiently large, there is an embedding $g: X\hookrightarrow \mathbb{R}^\ell$ which is unique up to isotopy. The map $f':X\to Y\times \mathbb{R}^\ell$ given by $f'(x)=(f(x),g(x))$ is then an embedding, and has a tubular neighbourhood $N$ as in Dan's answer. Note that the normal bundle $\nu_{f'}$ of $f'$ has dimension $\dim(Y)-\dim(X)+\ell$. There is a Pontrjagin-Thom collapse map $F:(Y\times\mathbb{R}^\ell)_+\to T(\nu_f')$, where $+$ denotes a one-point compactification. After making a standard identification, this gives a pointed map $F: \Sigma^\ell Y_+\to T(\nu_{f'})$, where $\Sigma^\ell$ denotes the $\ell$-fold reduced suspension.

Now we can define the pushforward as the composition $$ H^\ast(X)\to \widetilde{H}^{\ast+\dim(Y)-\dim(X)+\ell}(T(\nu_{f'}))\stackrel{F^\ast}{\to}\widetilde{H}^{\ast+\dim(Y)-\dim(X)+\ell}(\Sigma^\ell Y_+) \to H^{\ast+\dim(Y)-\dim(X)}(Y). $$ Here the first map is a Thom isomorphism and the last is a suspension isomorphism, and all coefficients are integers.

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Here is an extrinsic definition of the class of $Y$. The kernel of the natural map $$ H^*(X) \to H^*(X \setminus Y) $$ is a graded ideal in $H^*(X)$. The lowest degree that this ideal is non-zero in is precisely $codim(Y)$ and in this degree the image is a free $\mathbb{Z}$-module on one generator, which is $\pm[Y]$. This works if, e.g., $X \to Y$ is a proper embedding of smooth manifolds. I learned this in a paper of Feher and Rimanyi (See Definition 2.6 and the examples in 2.7).

To get the correct choice of a sign, at least for varieties, I would compute the degree (as in pushforward to a point) of the associated classes. Only one of these will be positive.

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1  
Some typos: the map $H^*(X) \to H^*(X \setminus Y)$ is not surjective, and the ideal you want to look at is the kernel of that map. Nice, but it gives you the class of $Y$ up to sign; is there a way to get the right sign? –  abx Dec 11 '13 at 5:53
    
thanks for catching this. in the $K$-theory of varieties $K^0(X) \to K^0(X-Y)$ is surjective (and the kernel is the image of $K^0(Y)$), whence my confusion. –  Andrew Dec 11 '13 at 6:46
    
That's really cool; thanks! –  Nicholas Proudfoot Dec 11 '13 at 16:41

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