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Let $M$ be a compact oriented $n$-manifold, and let $H^*(M)$ denote its cohomology ring with coefficients in $\mathbb{R}$.

Let's say that a graded subalgebra $K^\bullet \subset H^\bullet(M)$ is a Lagrangian subalgebra if we have an isomorphism of graded vector spaces with bilinear forms $$ H^\bullet(M) \simeq K^\bullet \oplus (K^{n-\bullet})^*$$ where the left hand side has the bilinear form coming from Poincare duality, and the right hand side has the sum of the tautological bilinear forms $$ \tau_i : K^i \oplus (K^{n-i})^* \times K^{n-i} \oplus (K^i)^* \to \mathbb{R}$$ given by $$ \tau_i((x,\phi), (y,\psi)) = \psi(x) + (-1)^{i(n-i)}\phi(y) $$

This terminology is meant to make an analogy with the case of a symplectic vector space $(V,\omega)$, where a subspace $L \subset V$ is Lagrangian if $(V,\omega) \simeq (L \oplus L^*, \tau)$ with $\tau$ the tautological alternating form.

The question is simply

  • When does $H^*(M)$ admit a Lagrangian subalgebra? Is there a simple criterion?

More generally I wonder if this notion has been studied under some other name, and if there is a structure theory classifying such subalgebras in cases when they do exist.

Examples

If $M=S^n$ then the trivial subalgebra $K = \mathbb{R}$ is the unique Lagrangian subalgebra.

If $M$ is an oriented surface of genus $g$, then using standard generators $\{ a_i, b_i \}_{i=1\ldots g}$ for $H^1(M)$, the algebra generated by $a_1, \ldots, a_g$ is a Lagrangian subalgebra. More generally one can take the algebra generated by any Lagrangian subspace of the symplectic vector space $H^1(M)$.

If $M = \mathbb{CP}^2$, or more generally if $M$ has dimension $2k$ and $H^k(M)$ is odd-dimensional, then there is no Lagrangian subalgebra. This is because a Lagrangian subalgebra would give an isomorphism $H^k(M)$ with the even-dimensional space $K^k \oplus (K^k)^*$.

Motivation

Lagrangian subalgebras seem to correspond to subcomplexes with interesting topological properties.

For example, suppose $\iota : X \hookrightarrow M$ is a subcomplex "representing" a Lagrangian subalgebra of $H^\bullet(M)$ in the sense that there is an exact sequence $$ 0 \to K^\bullet \hookrightarrow H^\bullet(M) \xrightarrow{\iota^*} H^\bullet(X) \to 0 $$ with $K^\bullet$ a Lagrangian subalgebra.

Suppose also that $X$ can be thickened to an embedded trivial $S^k$-bundle $Y \subset M$, for some $k \geq 0$. Then the cohomology ring of the complement $M \setminus Y$ satisfies Poincare duality in dimension $(n-k-1)$. (In this sense the complement of $Y$ "looks" like a bundle over the $(k+1)$-disk with compact fiber, at least cohomologically.)

In the case where $M = S^n$ and $X$ is a point, representing the trivial Lagrangian subalgebra, one recovers here the familiar homotopy equivalence between $S^n \setminus S^k$ and $S^{n-k-1}$.

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  • $\begingroup$ In the surface example you can think about Lagrangian subspaces as coming from writing the surface as the boundary of a solid g-holed torus and looking at the induced map on cohomology. $\endgroup$ – Qiaochu Yuan Feb 17 '16 at 22:52
  • $\begingroup$ You can find the "half lives/half dies" @QiaochuYuan said in math.cornell.edu/~hatcher/3M/3M.pdf $\endgroup$ – AHusain Feb 17 '16 at 23:13
  • $\begingroup$ Can you explain what you mean by a "subcomplex" in your motivation? $\endgroup$ – Chris McDaniel Feb 17 '16 at 23:36
  • $\begingroup$ @ChrisMcDaniel: I was thinking of a compact manifold with the structure of a finite CW complex, and of X as being a subcomplex of that. But for the topological facts I stated you just need X and Y to have regular neighborhoods in M. $\endgroup$ – David Dumas Feb 18 '16 at 4:47
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Let's assume $M$ is connected and $n$-dimensional. A subalgebra of $H(M)$ is Lagrangian if and only if its vector space dimension is one half that of $H(M)$ and $K^n=0$.

If $n=2q+1$ then there are always such subalgebras. One of them has $K^i=H^i(M)$ if $i$ is even and $K^i=0$ if $i$ is odd. Another has $K^i=H^i(M)$ if $i=0$ or $q<i<n$ and $K^i=0$ if $0<i<q+1$ or $i=n$.

If $n=2q$ with $q$ odd, then you can always make an example by choosing $K^q$ to be Lagrangian with respect to the alternating Poincare duality form on $H^q(M)$ and putting $K^i=H^i(M)$ if $i=0$ or $q<i<n$ and $K^i=0$ if $0<i<q$ or $i=n$.

If $n=4k$ then you can do the same thing if the symmetric Poincare duality form on $H^{2k}(M)$ has a "Lagrangian form" (i.e. if the signature of the manifold is zero), but there is no Lagrangian subalgebra if the signature is not zero.

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By definition, a closed oriented $n$-dimensional manifold $M$ has a twisted double structure if $M=N\cup_h-N$ for an orientation-preserving self-homeomorphism $h:\partial N \to \partial N$ of the boundary of an oriented $n$-dimensional manifold $N$. The graded subalgebra $K^{\bullet}=ker(H^{\bullet}(M)\to H^{\bullet}(N)) \subset H^{\bullet}(M)$ is a Lagrangian subalgebra of the asymmetric Poincare duality structure on $H^{\bullet}(M)$. In fact, using ${\mathbb Z}[\pi_1(M)]$-coefficients the converse is true: a closed oriented $n$-dimensional manifold $M$ has a twisted double structure if and only if there exists such an asymmetric Lagrangian subalgebra. The obstruction to the existence of a twisted double structure is the Quinn obstruction to the existence of an open book structure on $M$, which takes value in the asymmetric Witt group for $n$ even, and is 0 for $n$ odd. (As usual, one has to treat the cases $n>4$ and $n \leq 4$ separately and be particularly careful with the low-dimensional cases). See Chapters 29 and 30 of my 1998 Springer book "High-dimensional knot theory" http://www.maths.ed.ac.uk/~aar/books/knot.pdf for the algebraic surgery treatment of this obstruction, including the references to the relevant work of Milnor, Smale, Barden, Wall, Winkelnkemper, Quinn and T.Lawson. The book also includes an appendix by Winkelnkemper on the history and applications of open books.

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