Consider the following fragment from the book "Compact quantum groups and their representation categories" by Neshveyev-Tuset (p72, in section 2.5):

$\ \ \ $ Assume $\mathscr{C}$ is a category having all the properties of a strict $\scr C^*$-tensor category except existence of direct sums and subobjects. First complete it with respect to direct sums. For this, consider the category $\mathscr{C}'$ consisting of $n$-tuples $(U_1,\ldots,U_n)$ of objects in $\mathscr{C}$ for all $n\geqslant 1$. Morphisms are defined by $$\operatorname{Mor}\big((U_1,\ldots,U_n),(V_1,\ldots,V_m)\big)=\oplus_{i,j}\operatorname{Mor}(U_i,V_j).$$ Composition and involution are defined in the obvious way. Note that the norm is uniquely determined by the $\rm C^*$-condition $\|T\|^2=\|T^*T\|$, and for its existence we do need condition (ii) (c) in Definition 2.1.1 to be satisfied in $\mathscr{C}$. The tensor product of $(U_1,\ldots,U_n)$ and $(V_1,\ldots,V_m)$ is defined as the $nm$-tuple consisting of objects $U_i\otimes V_j$ ordered lexicographically. The category $\mathscr{C}'$ has direct sums: they can be defined by concatenation.

As I understand it, the morphism spaces between objects $(U_1, \dots, U_n)$ and $(V_1, \dots, V_m)$ just consist of matrices of morphisms between these objects, and composition and involution are defined as the natural matrix operations.

Tensoring morphisms corresponds to the formal Kronecker product of matrices.

What I do not understand however, is how these morphism spaces become Banach spaces (which is a requirement for these kinds of categories). There is a sentence in the text that mentions something about this: "Note that the norm .... to be satisfied in $\mathscr{C}$". However, I do not understand what the author is saying here.

Can someone clarify?

  • 1
    $\begingroup$ It might help to note that condition (ii)(c) from Defn 2.1.1 is that given $T:U\rightarrow V$ the morphism $T^*T$ in $\operatorname{Mor}(U)$ is positive. But right now I do not see the answer to your question. $\endgroup$ Commented Nov 25, 2022 at 12:55
  • $\begingroup$ @MatthewDaws I think this is similar to how one can show that $M_n(A)$ is a $C^*$-algebra. We have to realise $\operatorname{End}((U_1, \dots, U_n))$ as a unital $*$-subalgebra of some unital $C^*$-algebra and then show that it is closed in the induced norm. I'm trying some stuff with Hilbert modules for this, and I think I'm getting close (but perhaps I'm overcomplicating the issue). $\endgroup$
    – Andromeda
    Commented Nov 25, 2022 at 13:00

1 Answer 1


If you care about completeness, you just want to observe that the norm defined this way restricts to the original norms on the direct summands $\operatorname{Mor}(U_i, V_j)$. Since there are only finitely many summands, a sequence in $\bigoplus_{i,j} \operatorname{Mor}(U_i, V_j)$ is a Cauchy sequence if and only if the components in the summands are so.

The remark about Definition 2.1.1 is about checking the C$^*$-condition for the norm defined this way. You define $\|T\|$ as the square root of ‘the norm’ of $T^* T$, but how do you determine this?

One way to do is to treat $\bigoplus_{i, j} \operatorname{Mor}(U_i, U_j)$ as a right Hilbert C$^*$-module over the C$^*$-algebra $\bigoplus_i \operatorname{Mor}(U_i, U_i)$. (You want to use that condition to obtain a Hilbert module.) Then $T^* T$ is an adjointable endomorphism of this Hilbert C$^*$-module, hence you can take its operator norm.

Now, according to this convention, $\|T^* T\|$ should be defined as the square root of the operator norm of $(T^* T)^* (T^* T)$ acting on the same module. You can then use the fact that adjointable endomorphisms form a C$^*$-algebra with respect to the operator norm, hence you have $\|T\|^2 = \|T^* T\|$.

  • $\begingroup$ Thanks for the clear answer! A side question: If $\mathscr{C}$ has subobjects and we perform this construction to obtain the category $\mathscr{C}'$ by adding finite direct sums, then does $\mathscr{C}'$ also have subobjects? $\endgroup$
    – Andromeda
    Commented Nov 26, 2022 at 17:22
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    $\begingroup$ If the morphism sets are finite dimensional, yes it does, because any object will be a direct sum of simple ones. Otherwise no I think, for example you can consider the category $\mathcal C$ with two objects $0, U$ and the only nontrivial morphism space being $\operatorname{Mor}(U,U)=C(T^2)$. The subjects in $\mathcal C^\prime$ correspond to vector bundles over $T^2$. $\endgroup$ Commented Nov 26, 2022 at 19:01

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