# When can I assure that the representation theory of a PROP is faithful?

Recall that a PROP is a symmetric monoidal category whose object-set is freely $\otimes$-generated by a single object $x\in P$. A representation or algebra for a prop $P$ in a symmetric monoidal category $C$ is a symmetric monoidal functor $f: P \to C$ — i.e. it is an object $f(x)\in C$ and some morphisms between tensor powers of $f(x)$ that satisfy all the relations in $P$. I am interested in the $\mathbb Q$-linear versions of all of these: a $\mathbb Q$-linear prop $P$ is a $\mathbb Q$-linear symmetric monoidal category whose object-set is freely $\otimes$-generated by a single object $x\in P$, and a representation thereof in a $\mathbb Q$-linear category $C$ is a $\mathbb Q$-linear symmetric monoidal functor. I will henceforth leave implicit the word "$\mathbb Q$-linear". By $\mathrm{Vect}$ I mean the category of $\mathbb Q$-vector spaces.

A general question you can ask is the following. Suppose you are given a prop $P$, and some presentation of it in generators and relations. Suppose you write some expression in the generators — i.e. you pick some morphism in the prop. Suppose furthermore that for every representation $P \to \mathrm{Vect}$, this morphism evaluates to $0$. Does it follow that the morphism is $0$ in $P$? Put another way: Does every prop have a faithful representation in $\mathrm{Vect}$?

The answer, of course, is "NO!". An example: let $P$ be the prop generated by the relation that the braiding $x\otimes x \to x\otimes x$ is minus the identity. Then the only representation of $P$ in $\mathrm{Vect}$ is $x = 0$, and in particular the identity map $x \to x$ evaluates to $0$ in this representation. On the other hand, $P$ has a non-zero representation in the category of super vector spaces.

My question, then, is for (checkable) conditions on a prop $P$ to assure that it does, in fact, have faithful representations in $\mathrm{Vect}$.

For example, the prop that I happen to care about has a presentation in which it is generated by (at most) one morphism between any two objects, and the relations are all (homogeneous) linear and quadratic in the generators. I could imagine this to be the type of condition that might assure faithfulness of representations. I would like to know that my prop has a faithful $\mathrm{Vect}$-representation, because I can prove that in any $\mathrm{Vect}$-representation a certain morphism evaluates to $0$, by choosing a basis for the underlying vector space of the representation. Of course, this proof does not universalize, but maybe some other results assure me that the morphism is universally $0$.

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I asked the same thing on MO mathoverflow.net/questions/121838/…, so put my answer or rather trivial motivation to others here. If you look at problem of knot invariants, tqft construction and so on you noting that combinatorially this is the same thing as building representations of PROPs. For example you may consider category of tangles as operad with many outputs, and problem of finding full invariant of knots as building faithfull representation of such PROP. – Bad English Mar 1 '13 at 22:24
It is funny that from general point of view it is difficult to say why this operad is so complex. And why in PROPs there are no free-object, action on itself concept. But may be there exists some general statements. For example something like reduction to other operads. Let's back to knots for instance. What Drinfeld explanation of Kontsevich integral do - actually we move from the problem of building representations of tangle category to building their infinitesimal version - representations of chord diagrams, which much simpler. – Bad English Mar 1 '13 at 22:25
This is a nice question. Here is another example of a PRO with no faithful representations in Vect, that does not require linear structure. Let there be one generating object $A$ with a right dual, and morphisms $f:A\to A$ and $g: A \to A$ such that $f \circ g = \text{id}_A$. The right dual property ensures that for any representation $Z$ in Vect, $Z(A)$ is finite-dimensional, and then $Z(f) \circ Z(g) = \text{id}_{Z(A)}$ implies $Z(f) = Z(g)^{-1}$. But $f \neq g^{-1}$ in the PRO. – Jamie Vicary Jan 15 '15 at 20:30