A finite dimensional algebra $A$ over a field $K$ is called a Frobenius algebra in case there exists a $K$-linear map $f: A \rightarrow K$ such that $ker(f)$ contains no non-zero right ideal of $A$ (such an $f$ is called trace of $A$). $A$ is called symmetric in case additionally $f(xy)=f(yx)$ for all $x,y \in A$. $A$ is called weakly-symmetric in case every indecomposable projective module $P$ has the property that $top(P)=soc(P)$.
Symmetric implies weakly symmetric, which implies being Frobenius. But the other directions are not true in general.
Question: Is there a characterisation when a Frobenius algebra $A$ is weakly-symmetric using the trace map $f$?
Some experiments suggest that maybe in case $A$ is weakly-symmetric we have that $f(xy)=s_{x,y} f(yx)$ for some non-zero $s_{x,y} \in K$ for all $x,y$. Is this true?
In case this is true this would prove an old guess by me that being weakly symmetric over the field with two elements implies that the algebra is symmetric (since in this case the values of $f$ are either 0 or 1).
Note that there is the following connection between $f$ and the socles $soc(P)$ of the indecomposable projective modules $P$ in case $A=KQ/I$ is a quiver algebra:
$soc(e_i A)=<a_i>$ is 1-dimensional and spanned by a maximal path $a_i$ (I guess it can always be choosen to be a path, or could it be that it is not possible and it must be a sum of paths with coefficients?) and then $f(a_i)=1$ and $f(r)=0$ for all $r$ in a basis of $A$ that contains the elements $a_i$.
So at least for quiver algebras, the problem is very combinatorial.
