Dual Representation: Why Use g^{-1}?

  • Context: Graduate 
  • Thread starter Thread starter Pietjuh
  • Start date Start date
  • Tags Tags
    Dual Representation
Click For Summary

Discussion Overview

The discussion revolves around the concept of dual representations in representation theory, specifically addressing the use of the inverse element \( g^{-1} \) in the definition of the dual representation \( \rho^* \). Participants explore the implications of this definition and its relationship to the properties of representations, including the transformation of left representations to right representations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of using \( g^{-1} \) in the definition of the dual representation \( \rho^* \), suggesting a lack of clarity on this choice.
  • Another participant argues that using \( g \) instead of \( g^{-1} \) would result in an anti-homomorphism rather than a homomorphism, thus necessitating the use of \( g^{-1} \) to maintain the structure of a representation.
  • It is noted that taking duals interchanges the order of composition, which transforms a left representation into a right representation, and that groups possess an anti-involution that allows for correction back to a left representation.
  • A participant seeks clarification on whether the definition of the dual map is standard or derived from other principles.
  • Another participant describes how to define the action of a group \( G \) on the dual space, emphasizing the linear map's dependence on how it acts on vectors, and proposes a specific definition involving \( g^{-1} \) to correct the order of operations.
  • There is a mention of the relationship between matrix transposition and the properties of dual spaces, indicating that the reversal of order can be derived from basic linear algebra concepts.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of using \( g^{-1} \) in the definition of dual representations. While some agree on the need for this adjustment to maintain representation properties, others question the foundational reasoning behind it. The discussion remains unresolved regarding the clarity and derivation of the dual map definition.

Contextual Notes

Participants reference various assumptions about the nature of representations and the properties of groups, including the distinction between left and right representations. There is an acknowledgment of specific cases where the general statements may not hold, such as when the group or representation is abelian.

Pietjuh
Messages
75
Reaction score
0
I've been starting to study some things about representation theory. I've come to the point where they introduced the dual of a representation.

Suppose that [itex]\rho[/itex] is a representation on a vector space V.
They then define the dual representation [itex]\rho^*[/itex] as:

[tex]\rho^*(g) = \rho(g^{-1})^t: V^* \to V^*[/tex]

But the thing is that I don't see why they use [itex]g^{-1}[/itex] in this definition instead of just g?
 
Physics news on Phys.org
Because otherwise it would not be a representation (the map would not be a homomorphism, but an anti-homomorphism, that is denoting your notional maps as f, f(gh)=f(h)f(g))
Taking duals interchanges the order of composition, ie it makes a left representation into a right representation, fortunately groups possesses this anti-involution that makes you able to correct it and turn it into a left representation again. Left means make the matrix act on the left, right means make the matrix act on the right.

Notice they *do* actually use g to define the representation p*, ie they do tell you how to work out p*(g), and it is p(g) 'inverse transpose'.

Given some representation p, there are many things that we can do to get another representation. This is just one of them, and it so happens the vector space of the representation is V*.

You shuold check that you do indeed find that the representation

f(g)=p(g) transpose

is not generically (which means 'usually', or 'except for certain cases' such as G being abelian, or p(G) being abelian) a (left) representation, ie the map f is not a group homomorphism.
 
Last edited:
matt grime said:
Taking duals interchanges the order of composition, ie it makes a left representation into a right representation, fortunately groups possesses this anti-involution that makes you able to correct it and turn it into a left representation again. Left means make the matrix act on the left, right means make the matrix act on the right.

Is this the definition of a dual map? Or is it derivable from something else?
 
Given a representation, p,V, how can you make G act on the dual space? if f is in the dual space then the *only* obvious action of G on f is to define

g.f(v)=f(g.v)
for all v in V.: remember it suffices to define a linear map by how it acts on vectors, so we defince g.f to be the linear map that sends v go f(g.v).

All this is saying is that End(V) maps to End(V*) by taking transposes. And and (AB)^T = (B^T)(A^T) which we all learned in our first lecture on dual spaces. So it's derivable just from elementary linear/quadratic maths.Let's prove it reverses the order (this is just revision but in a different notation, probably): if you do this (gh).f(v) = f((gh).v)=f(g(h.v))=g.f(h.v)=h.(g.f(v), so it naturally changes the order.

But there is a way to correct this for *group representations*, by making

g.f(v)=f(g^{-1}(v))

for all v.

The composition of inverting matrices and then transposing them swaps the composition over twice.
 
Last edited:

Similar threads

  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
10K