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