Cohomology class of the diagonal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:04:45Z http://mathoverflow.net/feeds/question/74181 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74181/cohomology-class-of-the-diagonal Cohomology class of the diagonal Descartes 2011-08-31T17:40:05Z 2011-09-01T05:29:04Z <p>I have a pretty basic question:</p> <p>given is a complex manifold of dimension n or a smooth projective variety over $k$ (char 0, algebraically closed). Then people are often speaking of "the cohomology class of the diagonal". Can you explain me how you realize this class, i.e. in which cohomology group does it land, how do you formally construct it, etc.</p> <p>Thank you!</p> http://mathoverflow.net/questions/74181/cohomology-class-of-the-diagonal/74196#74196 Answer by Neil Strickland for Cohomology class of the diagonal Neil Strickland 2011-08-31T20:02:00Z 2011-08-31T20:02:00Z <p>I'll answer for smooth closed manifolds. I'll also assume choose Riemannian metrics everywhere. (The space of such metrics is convex and therefore contractible, so anything depending on the metric will usually be canonical up to homotopy.) Suppose we have a smooth embedding $f:N\to M$ of codimension $d$. This gives a $d$-dimensional vector bundle $\nu_f$ over $N$, whose fibre at $x$ is the orthogonal complement to $f_*(T_xN)$ in $T_{f(x)}M$. Next, for $(x,v)\in E(\nu_f)$ I define $g_1(x,v)$ to be $\alpha(1)$, where $\alpha:\mathbb{R}\to M$ is the geodesic with $\alpha(0)=f(x)$ and $\dot{\alpha}(0)=v$. This map $g_1:E(\nu_f)\to M$ will be an embedding on some neighbourhood of the zero section. Choose a smooth embedding $h:E(\nu_f)\to E(\nu_f)$ whose image is close to the zero section, and which is the identity on some even smaller neighbourhood of the zero section. Now put $g=g_1\circ h:E(\nu_f)\to M$; this is called a <em>tubular neighbourhood</em> of $f$. We can define a map $f^!:M_\infty\to E(\nu_f)_\infty$ of one-point compactifications by $f^!(g(x,v))=(x,v)$ and $f^!(m)=\infty$ whenever $m$ is not in the image of $g$. This is called the <em>Pontrjagin-Thom construction</em>. This in turn gives a map $\widetilde{H}^*(E(\nu_f)_\infty)\to\widetilde{H}^*(M_\infty)$. Now $M$ is compact, so the point $\infty\in M_\infty$ is isolated, so $\widetilde{H}^*(M_\infty)=H^*(M)$. On the other hand, $E(\nu_f)_\infty$ is otherwise known as the Thom space $N^{\nu_f}$, so (provided that $\nu_f$ is oriented) there is a canonical element $u\in \widetilde{H}^d(N^{\nu_f})$ (called the <em>Thom class</em>) such that $\widetilde{H}^d(N^{\nu_f})$ is freely generated by $u$ as a module over $H^*(N)$. We now have an element <code>$v=(f^!)^*(u)\in H^d(M)$</code>; this is called the cohomology class of $N$ (or of $f$) in $M$.</p> <p>The cohomology class of the diagonal is obtained by applying this procedure to the diagonal embedding $f:M\to M^2$ defined by $f(m)=(m,m)$. </p> http://mathoverflow.net/questions/74181/cohomology-class-of-the-diagonal/74199#74199 Answer by Kevin Lin for Cohomology class of the diagonal Kevin Lin 2011-08-31T20:21:28Z 2011-08-31T20:58:07Z <p>Let $\Delta : M \to M \times M$ be the diagonal map. Since $M$ is a complex manifold, say of complex dimension $n$, it has a canonical orientation class $[M] \in H_{2n}(M, \mathbb{Z})$. Then you can take the pushforward in homology to get $\Delta_\ast [M] \in H_{2n}(M \times M, \mathbb{Z})$. If $M$ is compact then $M \times M$ is also compact and you can use Poincare duality to get an element in $H^{2n}(M \times M, \mathbb{Z})$. This is the cohomology class of the diagonal.</p> <p>More generally, the words to look up are Thom isomorphism theorem or Gysin sequence or Gysin map. The inclusion $\Delta$ induces a Gysin map $\Delta_\ast: H^i(M) \to H^{i-(-2n)}(M \times M)$. The cohomology class of the diagonal is the image of $1 \in H^0(M)$ under this map. You can do the same kind of thing in $K$-theory, Chow groups, etc.</p> http://mathoverflow.net/questions/74181/cohomology-class-of-the-diagonal/74222#74222 Answer by anonymous for Cohomology class of the diagonal anonymous 2011-09-01T03:11:01Z 2011-09-01T05:29:04Z <p>Here is one answer for the question of Descartes in the case $X$ is compact Kahler manifold and the coefficient is $\mathbb{Q}$: Then we can work with coefficient $\mathbb{C}$ as well. This answers uses the de Rham cohomology. </p> <p>If $X$ is a compact Kahler manifold of complex dimension $n$, then <code>$\Delta_X\subset X\times X$</code> is a cycle of (real) dimension $2n$, hence acts on smooth $2n$ forms on $X\times X$ by integration. Also, Delta has no boundary, hence it is closed (by Stokes theorem). Thus it represents a homology class in <code>$H_{2n}(X\times X)$</code>. Now, <code>$\Delta_X$</code> acts on a form by integration on the restriction of that form on <code>$\Delta_X$</code>. Since <code>$\Delta_X$</code> is a complex manifold of dimension $n$, if $\varphi$ is a $(p,q)$ form on $X\times X$ with $p+q=2n$, its restriction to <code>$\Delta_X$</code> is nonzero iff $p=q=n$. This shows that <code>$\Delta_X$</code> is in <code>$H_{n,n}(X)$</code>, and so its Poincare dual represents a cohomology class in $H^{n,n}(X)$. </p>