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 $S^{n-k-1}$.



"Lagrangian" subalgebra of cohomology, with respect to Poincare duality?
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. 


