The String Coffee Table

All I really do in the following is to take some elementary linear algebra reasoning and use it within the context of 2-vector spaces with 2-scalar product that I discussed in part I ($\to$). Might be slightly boring, but deserves to be said.

So, to recapitulate, we have a braided tensor category $C$ with duals, we call (right) $C$-module categories 2-vector spaces and tend to restrict attention to $C$-modules that are $C$-internal left $A$-module categories $\mathbf{V} = \multiscripts{_A}{\mathrm{Mod}}{}$.

We assume we have an internal $\mathrm{Hom}$ functor on these 2-vector spaces $$\mathrm{IHom} : \mathbf{V}^\mathrm{op} \times \mathbf{V} \to C$$ which we regard as the 2-scalar product on $\mathbf{V}$.

Moreover, for the time being I restrict attention to those situations (which are relevant for 2D TQFT and RCFT) where the internal $\mathrm{Hom}$ is given by $$\langle \vec v ,\vec w \rangle := \mathrm{IHom}(\mathbf{V},\mathbf{W}) = \vec v^\vee \otimes_A \vec w \,,$$ where $\vec v, \vec w \in \mathrm{Obj}(\mathbf{V})$ and where $(.)^\vee$ is the operation on bimodules ($\vec v$ is to be regarded as an $A$-$1$-bimodule) induced by the duality operation in $C$.

$A$-$A$ bimodules $B$ act as 2-linear maps (functors of $C$-module categories) on $\mathbf{V}$. Their funcorial adjoint coincides with their adjoint with respect to $\langle\cdot,\cdot,\rangle$ and is given by $$B^\dagger := B^\vee \,.$$

We are interested in understanding when a 2-endomorphism $B : \mathbf{V} \to \mathbf{V}$ admits something like a basis of eigen-2-vectors.

So the first thing to clarify is what we want to understand under

$\bullet$ an eigen-2-vector,

$\bullet$ a basis of 2-vectors.

There is probably some room for being very sophisticated at this point. Not the least because there is some reason to suspect that all this can yield anything really interesting only when our 2-vector spaces are categories that are triangulated ($\to$), like derived categories are. That’s because the triangulation property can be understood as a being realization of 2-subtraction. (Recall ($\to$) that in 2D QFT 2-vectors correspond to D-branes, and triangles encode brane/anti-brane annihilation processes ($\to$) - subtraction of branes).

But I won’t be quite that sophisticated just yet. I want to get some basic things straight in a simpler context, to see how to properly generalize from there on.

With this disclaimer out of the way, I’ll state the notion of eigen-2-vector and of basis of eigen-2-vectors that I shall consider here.

Definition 1) Given a $C$-linear 2-endomorphisms $$B : \mathbf{V} \to \mathbf{V}$$ we say that a 2-vector $\vec v \in \mathrm{Obj}(\mathbf{V})$ is a 2-eigenvector of $B$ iff there is $E \in \mathrm{Obj}(C)$ such that $$B \otimes_A \vec v \simeq \vec v \otimes E \,.$$ More precisely, we should give a more diagrammatic definition. A 2-eigenvector of $B$ with eigen-2-value $E \in \mathrm{Obj}(C)$ is a morphism $$C \overset{\vec v}{\to} \mathbf{V}$$ of internal right $1$-modules (1 is the tensor unit in $C$), such that $$\array{ C & \overset{\vec v}{\to}& \mathbf{V} \\ \vec v\downarrow\;\; &\overset{\sim}{\Leftarrow}& \;\;\;\;\;\;\;\;\;\;\;\downarrow B\otimes_A \cdot \\ \mathbf{V} & \overset{\cdot \otimes E}{\to} & \mathbf{V} } \,.$$

Definition 2) An orthonormal 2-basis of $\mathbf{V}$ is a collection $(\vec e_i)_i$ of 2-vectors such that every object of $\mathbf{V}$ is isomorphic to $\oplus_i \vec e_i \otimes c_i$ for some $c_i \in C$, and such that $$\langle \vec e_i , \vec e_j \rangle$$ is isomorphic to the tensor unit in $C$ if $i=j$ and isomorphic to the 0-object in $C$ otherwise.

There is clearly room for improvement in these definitions, but right now all I care about is that this way we can state the following obvious

Definition 3) An orthonormal 2-basis $(\vec e_i)_i$ is called a eigen-2-basis of $B : \mathbf{V} \to \mathbf{V}$ iff $$B \simeq \oplus_i\; \vec e_i \otimes E_i \otimes \vec e_i^\vee$$ for some $E_i \in \mathrm{Obj}(C)$.

The fact that all our fancy 2-linear algebra here follows - up to some isomorphisms - the logic of ordinary linear algebra makes the following observation a triviality - but a possibly interesting one:

Observation. If $B$ admits a basis of eigenvectors then $$B \otimes_A B^\dagger \simeq B^\dagger \otimes_A B \,.$$

Proof. We have $$\begin{aligned} B \otimes_A B^\dagger &\overset{def 3}{=} \left(\oplus_i\; \vec e_i \otimes E_i \otimes \vec e_i^\vee \right) \otimes_A \left(\oplus_i\; \vec e_i \otimes E_i^* \otimes \vec e_i^\vee \right) \\ &\overset{\text{def. 2}}{\simeq} \left(\oplus_i\; \vec e_i \otimes E_i \otimes E_i^*\otimes \vec e_i^\vee \right) \end{aligned}$$ as well as $$\begin{aligned} B^\dagger \otimes_A B &\overset{def 3}{=} \left(\oplus_i\; \vec e_i \otimes E_i^* \otimes \vec e_i^\vee \right) \otimes_A \left(\oplus_i\; \vec e_i \otimes E_i \otimes \vec e_i^\vee \right) \\ &\overset{\text{def. 2}}{\simeq} \left(\oplus_i\; \vec e_i \otimes E_i \otimes E_i^*\otimes \vec e_i^\vee \right) \end{aligned} \,.$$ The isomorphism follows because we assume $C$ to be braided. $\square$

This is essentially the condition I talked about last time.

All right, nothing overly exciting has happened here. Somebody should try to figure out

$\bullet$ what happens when we weaken the above condition to $$\mathrm{Tr}(B \otimes_A B^\dagger) \simeq \mathrm{Tr}(B^\dagger \otimes_A B) \,,$$ where $\mathrm{Tr}$ is the Kaparanov-Ganter 2-trace;

$\bullet$ how this condition relates to the somewhat reminiscent condition on duality defects in RCFT (see the discussion in the previous entry).