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$\require{AMScd}$ Let $\mathcal{C}$ be a tensor category. An object $X$ in $\mathcal{C}$ will be called selfdual if $X^* = X$. Let $A,B,M$ be selfdual objects in $\mathcal{C}$. Let $i_X: X \to M$, with $X=A,B$, be two monomorphisms, and consider their pullback below:

$$\begin{CD} P @>j_A>> A\\ @VVj_BV @VVi_AV\\ B @>i_B>> M \end{CD}$$

Question: Is the object $P$ also selfdual (up to isomorphism)?

Following Freyd's book [Fr64, pages 19, 37-40], $i_X$ is a (equivalent class representative) subobject of $M$, and $P$ is the intersection $A \cap B$. If we can prove something like $(A \cap B)^* \simeq A^* \cap B^*$, then we are done.

If required, we can assume, for $X=A,B$, that $i_X^* \circ i_X = {\rm id}_X$, implying that $i_X$ is a split monomorphism. Thus, by splitting lemma, we can assume without loss of generality that $$M = X \oplus M/X$$ with $i_X = {\rm id}_X \oplus 0$ (recall that $X = X \oplus 0$).

The dual of above diagram is the following pushout of epimorphisms $i_X^*: M \to X$.

$$\begin{CD} M @>i_A^*>> A\\ @VVi_B^*V @VVj_A^*V\\ B @>j_B^*>> P^* \end{CD}$$

Semisimple case: Let $(X_i)$ be the simple objects up to isomorphism. Without loss of generality, we can take $M=\bigoplus_i M_i \otimes X_i$, $A=\bigoplus_i A_i \otimes X_i$ and $B=\bigoplus_i B_i \otimes X_i$, where $A_i$ and $B_i$ are subspaces of the multiplicity space $M_i$, for all $i$. Without loss of generality, we can take $i_A$ and $i_B$ induced by the inclusions $A_i, B_i \subset M_i$. Then $P=\bigoplus_i P_i \otimes X_i$, with $P_i = A_i \cap B_i$. But $A$, $B$, $M$ are selfdual, so $M_{i^*} = M_i$, $A_{i^*} = A_i$ and $B_{i^*} = B_i$. Thus $$P_{i^*} = (A_i \cap B_i)^* = A_{i^*} \cap B_{i^*} = A_i \cap B_i = P_i,$$ for all $i$, meaning that $P^* = P$.


Reference:

[Fr64] Freyd, Peter. Abelian categories. An introduction to the theory of functors. Harper's Series in Modern Mathematics. Harper & Row, Publishers, New York, 1964. {\rm xi}+164 pp.

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  • $\begingroup$ Certainly this should be the case for fusion categories (or more generally, semi-simple tensor categories), since then P is a subobject of the sum of A and B. $\endgroup$
    – Eric S.
    Commented Aug 22 at 20:33
  • $\begingroup$ The proof for the semisimple case has now been included. $\endgroup$ Commented Sep 4 at 13:40

1 Answer 1

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This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

See the footnote at the bottom of this post for the notation.

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $SL(2,3)$ act irreducibly on $N$ as the binary tetrahedral group. Then $SL(2,3)$ acts on $E$, and we can form the semidirect product $G=E\rtimes SL(2,3)$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

because $SL(2,3)$ acts on $N$ with determinant one, and we can then do the same construction as above.

Footnote: Notation for modules. The OP has asked me to explain the notation for modules.

When I write

$$\begin{matrix}A\\B\\C\end{matrix}$$

what I mean is a uniserial module with composition factors $A$, $B$, $C$. So there is a unique submodule, isomorphic to $C$; modulo this there is a unique submodule, isomorphic to $B$; and modulo this, it's the simple module $A$.

The structure of the projective cover of the trivial module for $A_5$ given above is that there is a unique top composition factor, $k$, and a unique bottom composition factor, also $k$, and the radical modulo the socle is a direct sum of two uniserial modules. Similarly, the second self dual submodule of this has a unique top and unique bottom composition factors isomorphic to $k$, and the radical modulo the socle is $M\oplus N$.

There is a theory of diagrams for modules, developed in my paper with Carlson, "Diagrammatic methods for modular representation theory and cohomology", of which this is part. Beware though that not all modules have a nice diagram of this form.

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  • $\begingroup$ Is it also a counter-example for the split case as I specified? $\endgroup$ Commented Sep 5 at 10:46
  • $\begingroup$ A counterexample for the same group exists for the split case. I'll add it to the answer. $\endgroup$ Commented Sep 5 at 10:51
  • $\begingroup$ Do you expect the existence of such counter-examples over $\mathbb{C}$ also? $\endgroup$ Commented Sep 5 at 11:27
  • $\begingroup$ Yes, there's a supersymmetric example of the same sort. I'll make yet another edit to the answer. $\endgroup$ Commented Sep 5 at 11:42
  • $\begingroup$ I changed it from $S_3$ to $SL(2,3)$ because I realised that $S_3$ does not act with determinant one. The effect of this was that the socle was the sign representation, and the module was therefore not self-dual. It now is. $\endgroup$ Commented Sep 5 at 14:50

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