Is this the appropriate specific generalisation of twin room


I am currently considering a framework that I am attempting to select- my approach being to draw things up right into the best feasible generalization (based upon the little bits I'm certain concerning) and also slim it below there.

The scenario I have is rather comparable to the twin room framework in vector rooms, though likely much less mannerly and also vector rooms alone will certainly not suffice.

The Construction

Take into consideration a vector room $V$ over an area $k$- its twin room $V^*$ shows up normally as the set linear of maps

$$w^* : V \to k$$

which, together, create a vector room in themselves. We can popularize this to approximate groups $A$, $B$, $C$ by establishing $B=hom(A,C)$. After that, at the very least in some feeling, $B=A^*$. Until now so typical, yet I desire extra: a wonderful building of twin rooms is that a component of $V \otimes V^*$ can be canonically viewed as a component of $hom(V,V)$- this is as a result of the manner in which $k$ acts upon $V$ by reproduction. We can resemble this by allowing $C$ be a monoid acting upon $A$.

In recap: A group $A$ acted upon by a monoid $C$ and also a twin A $A^*:=hom(A,C)$

I am specifically curious about when $A$ is additionally a monoid, specifically so when $A$ is a room of stochastic matrices.


So this isn't also not likely a building and construction, actually it's possibly forehead-slappingly popular, so:

  • What is it called, if anything?
  • In what instances can we have $A=B=C$? Is that always a permutation team as an example?
  • Exist any kind of valuable approved instances, besides vector rooms?
  • Even better, theories ??? Papers???

As you can possibly inform, I am no group philosopher, so any kind of aid would certainly be outstanding.

2019-05-07 07:37:19
Source Share
Answers: 2

I think the appropriate generalization of twin rooms is that of a dual object in a tensor category, which I will certainly think symmetrical for ease.

Remember what makes a twin room of a vector room job : We have a map $V \times V^* \to k$ (for $k$ the ground area). The trouble is, this isn't a homomorphism in the group of vector rooms ; it is instead a bilinear map. So you can consider it as a map $V \otimes V^* \to k$ rather. This is why you require a tensor framework to consider duals.

This isn't sufficient, however, due to the fact that we require to recognize that the pairing is nondegenerate. One means to share this is that there's a map $k \to V \times V^*$ mapping 1 to the "Casimir component" (which is the amount $\sum e_i \otimes e_i^{\vee}$ where $e_i$ varies over a basis of $V$ and also $e_i^{\vee}$ the twin basis ; it is independent of the selection of $e_i$ as a fast calculation programs). The Casimir morphism pleases the problem that $V \to (k) \otimes V \to (V \otimes V^*) \otimes V \to V \otimes (V^* \otimes V)$ is simply the identification.
Alternatively, this suffices to show that the pairing is nondegenerate.

So, anyhow, just how does this make good sense in a symmetrical tensor group? Primarily, $V$ is the object, $V^*$ the presumptive twin, and also $k$ changed by the unital object. This definition is totally arrowhead - logical, and also all of it experiences customarily. It is a workout to examine that the twin is one-of-a-kind.

Some instances:

  1. This accompanies the common twin in the group of vector rooms

  2. This accompanies the twin sheaf if one is operating in the group of in your area free sheaves on a system

  3. This represents the twin (contragredient) depiction in the (tensor) group of depictions of any kind of Hopf algebra (so this consists of depictions of limited teams and also Lie algebras)

Oh, and also what takes place if you do not have a symmetrical tensor group? After that you need to bother with "left" and also "appropriate" duals, specifically. For even more concerning all this, I advise the notes of Pavel Etingof on tensor groups.

2019-05-08 23:47:47

You can review duals in a monoidal group (which might not be symmetrical). This has actually been stated by Akhil.

Allow $V$ and also $W$ be things (in your monoidal group) with $K$ the identification for tensor item. After that you call for morphisms $K\rightarrow V\otimes W$ and also $W\otimes V\rightarrow K$ which please the zig-zag identifications (supposed due to the fact that this comes to be clear if you attract string layouts).

Officially the zig-zag identifications are $$V=K\otimes V\rightarrow V\otimes W\otimes V\rightarrow V\otimes K=V$$ is the identification map and also $$W=W\otimes K\rightarrow W\otimes V\otimes W\rightarrow K\otimes W=k$$ is the identification map.

This amounts $Hom(W\otimes X,Y)=Hom(X,V\otimes Y)$ and also $Hom(X\otimes V,Y)=Hom(V,Y\otimes W)$ (both all-natural in $X$ and also in $Y$.

After that you claim $V$ is left/right twin to $W$ and also $W$ is right/left twin to $V$ ( I can never ever bear in mind which). After that we can specify $V$ to be twin to $W$ if it is left twin and also appropriate twin.

2019-05-08 04:21:50