Do Frobenius subalgebras form a lattice?

Question

A finite-dimensional, unital, associative algebra $A$ over a field $k$ is termed a Frobenius algebra if it is endowed with a nondegenerate bilinear form $\sigma : A \times A \to k$ satisfying the equation $\sigma(a \cdot b, c) = \sigma(a, b \cdot c)$ for all elements $a, b, c \in A$.

A Frobenius subalgebra $B$ of $A$ is defined as a unital subalgebra for which the restriction of $\sigma$ to $B \times B$ remains nondegenerate.

Our investigation focuses on determining whether the poset of Frobenius subalgebras of $A$ can be endowed with a lattice structure. Specifically, we seek to ascertain whether every pair of Frobenius subalgebras of $A$ possesses a unique supremum (join) and a unique infimum (meet) within this poset framework.

Dave Benson provided an example illustrating Frobenius subalgebras whose intersection does not result in a Frobenius subalgebra. Consequently, the meet, if it exists, is not necessarily defined by the intersection.

Answer 1

No. Let $A$ be the cohomology algebra of a Riemann surface of genus $g \geq 3$, i.e. it has a basis of the form $1, x_1,\dots, x_{2g}, z$ with $1$ the unit, $x_i z=0=zx_i $ for all $i$, $$x_i x_j = \begin{cases} z & \textrm{if } j= i+g \\ -z & \textrm{if } i=j+g \\ 0 & \textrm{otherwise}\end{cases}$$ and $\sigma(a \cdot b)$ extracting the coefficient of $z$ in $ab$.

Then $\langle 1, z\rangle$ is a Frobenius subalgebra, so the subset of the poset consisting of all subalgebras containing it is the set of elements of the poset $\geq$ a fixed element and therefore is a lattice if the original poset is a lattice. So it suffices to show that this poset is not a lattice.

Each subalgebra containing $1,z$ is uniquely determined by a subspace of the vector space generated by $x_1,\dots, x_{2g}$, and every subspace gives a subalgebra. However, a subspace is a Frobenius algebra if and only if the symplectic bilinear form is nondegenerate on it.

So it suffices to check that the subspaces of a symplectic vector space of dimension $2g \geq 6$ where the restriction of the symplectic bilinear form is nondegenerate do not form a lattice.

To see this, consider the subspaces generated by $x_1, x_{g+1}$ and $x_1, x_{g+1} + x_2$. If the subspaces where the form is nondegenerate form a lattice, then there exists a minimum subspace where the form is nondegenerate containing both, which must have dimension at least $4$. But $x_1, x_2 , x_{g+1},x_{g+2}$ and $x_1,x_2, x_{g+1} , x_{g+2} +x_{g+3}$ are two different subspaces of dimension $4$ containing both, so no $4$-dimensional subspace is minimal since none can be contained in both.