# Math Help - Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

1. ## Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

I am reading the first set of examples in Chapter 18 Representation Theory & Character Theory and Dummit and Foote - and cross referencing with examples in James and Liebeck's "Representations and Characters of Groups"

I have a (possible) problem with Example 2 page 844 from D&F and am hopign someone can help!

The text of the problem in D&F is as follows: (see attachement of page 844 D&F)

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Let V = FG and consider this ring as a left module over itself. Then V affords a representation of G of degree equal to |G|. If we take elements of G as a basis for V, then each $g \in G$ permutes these basis elements under the left regular representation:

$g \cdot g_i = g g_i$

[Note/question by Peter: I thought we had to specify an action of a ring element g on an element v of V= FG - but then I suppose if we specify an action on each basis element then this approximates to the same thing??]

With respect to this basis of V, the matrix of the group element g has a 1 in row i and column j if $g g_i$ , and has 0s in alll other positions.

This linear or matrix representation is called the regular representation of G

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Since I understand maths better when I have a concrete example I took an example from James and Liebeck - see attachement pages 56 -57 of J&L)

J&L took V = FG where $G = C_3$, the cyclic group of order 3 $< a: a^3 = e >$ where the set of elements $\{ g_1, g_2, g_3 \} = \{ e, a, a^2 \}$

Now J&L use a slightly different approach and notation so I tried to follow the D&F approach and notation and work through the problem, thus ....

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Consider $g = g_1 = e$

$g g_1 = ee = e$ so $g g_1 = g_1$

Thus we have 1 in row 1 and column 1

$g g_2 = ea = a$ so $g g_2 = g_2$

Thus we have 1 in row 2, column 2

$g g_3 = e a^2 = a^2$ so $g g_3 = g_3$

Thus, again following D&F we have a 1 in row 3 and column 2

Now apart from the 1s above the reset of the matrix for $g = g_1 = e$ is the identity matrix

Thus the linear mapping for $g_1 = e$ is as follows:

$e \longrightarrow \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{ar ray}\right)$

Mapping is $e \longrightarrow$

1 0 0
0 1 0
0 0 1

[This squares with J&L page 57 !

Consider now $g = g_2 = a$

$g g_1 = ae = a$ so $g g_1 = g_2$

Thus we have a 1 in row 2 and column 1

$g g_2 = a.a = e$ so $g g_2 = g_1$

Thus we have a 1 in row 1, column 2

$g g_3 = a a^2 = a$ so $g g_3 = g_2$

Thus we have a 1 in row 2 and column 3

Thus the linear mapping for $g_2 = a$ is as follows:

Mapping is $a \longrightarrow$

0 1 0
1 0 1
0 0 0

[PROBLEM - this does not sqare with J&L page 57 which gives

Mapping is $a \longrightarrow$

0 1 0
0 0 1
1 0 0 ]

Consider now $g = g_3 = a^2$

$g g_1 = a^2 e = a^2$ so $g g_1 = g_3$

Thus we have a 1 in row 3 and column 1

$g g_2 = a^2 a = e$ so $g g_2 = g_1$

Thus we have a 1 in row 1 and column 2

$g g_3 = a^2 a^2 = a$ so $g g_3 = g_2$

Thus we have a 1 in row 2 and column 3

Thus the linear mapping for $g_3 = a^2$ is as follows:

Mapping is $a^2 \longrightarrow$

0 1 0
0 0 1
1 0 0

[PROBLEM - this does not sqare with J&L page 57 which gives

Mapping is $a^2 \longrightarrow$

0 0 1
1 0 0
0 1 0 ]

1. Confirming that my general analysis following D&F is correct (or not!)

2. Resolving the problem between my working and that of J&L pages 56 -57 (please see attachement) [Hmmm ... is there some problem with rows and columns in the two approaches?]

Peter

2. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

the problem is, you've got the action wrong for how G acts on {e,a,a2}.

G acts "cyclically" on these elements, for example a.e = a, a.a = a2, and a.a2 = e. so the 3x3 matrix corresponding to a should be:

[0 1 0]
[0 0 1]
[1 0 0]

it's easier to understand this if you view G as the 3-element subgroup of S3 given by:

{e, (1 2 3), (1 3 2)}, via Cayley's theorem (here "a" corresponds to the 3-cycle 1-->2-->3-->1).

for some reason, you took a.a to be e, but a is of order 3, not 2.

let's look at a "bigger example", the group D3 of order 6. i'll write the basis elements of the group ring Q[D3] as {1,r,r2,s,rs,r2s}.

clearly 1 maps to the 6x6 identity matrix, that's rather boring. so let's consider (left)-multiplication by r:

1-->r
r-->r2
r2-->1
s-->rs
rs-->r2s
r2s-->s, which gives the 6x6 matrix:

[0 1 0 0 0 0]
[0 0 1 0 0 0]
[1 0 0 0 0 0]
[0 0 0 0 1 0]
[0 0 0 0 0 1]
[0 0 0 1 0 0].

note that we can write this in block form as:

[A 0]
[0 A], where A is the 3x3 matrix we found for a in C3 above. this is no coincidence-we get this nice decomposition because <r> = C3 is a normal subgroup of D3.

similarly, for r2, we get the matrix:

[0 0 1 0 0 0]
[1 0 0 0 0 0]
[0 1 0 0 0 0]
[0 0 0 0 0 1]
[0 0 0 1 0 0]
[0 0 0 0 1 0].

look at how the reflections "mess this up", for s we get the matrix:

[0 0 0 1 0 0]
[0 0 0 0 0 1]
[0 0 0 0 1 0]
[1 0 0 0 0 0]
[0 0 1 0 0 0]
[0 1 0 0 0 0], we still have a block decomposition, but the "blocks" are in the wrong place.

i leave it to you to guess what the other 2 matrices look like, and to show this set of 6 matrices is indeed closed under matrix multiplication (hint: use their "block form").

the fact that we can reduce this group representation to 3x3 blocks might lead you to suspect that using a 6-dimensional space may not be the lowest possible dimension, and you'd be right.

EDIT: it appears that actually D&F would take the TRANSPOSE of these matrices, while J&L would use these (because of the way the different authors define the mappings).

3. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

Yes, see my silly error ... no idea why I did that ... will correct that

Not quite sure why the action I took was wrong though ..

I am looking again at D&F and not they say that $g \cdot g_i = g g_i$ is the left regular representation ... and not the action ... I am assuming that this was my fundamental error (as distict from the silly error above! :-(

Peter

4. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

Just wondering how you would formally express the action?

Peter

5. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

which action are you speaking of? if |G| = n, we can take V to be Fn, and consider the action of G on V, in the very same way we regard G as a permutation (via left-multiplication) on its underlying set of elements. that is, if ggk = gjk, we consider the element σ of Sn that sends k to jk (as k runs from 1 to n) and say φ(g) is the F-linear map that maps ek (= gk, if we pick our basis elements to be the elements of G) to eσ(k).

6. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

Hmm ... so ... (tentatively) ... $g g_k = g_j$ is the action of G on the basis elements of V - and hence does specify the action of a ring element g on an element of V?

Is that correct?

Peter

7. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

Thanks ... will work through your extended example

Peter

8. ## Re: Regular Representations - Ch 18 Dummit and Foote & Ch 1-6 James & Liebeck

Originally Posted by Bernhard
Hmm ... so ... (tentatively) ... $g g_k = g_j$ is the action of G on the basis elements of V - and hence does specify the action of a ring element g on an element of V?

Is that correct?

Peter
remember that the ring elements are linear combinations of the group elements. so to define the ring action, just take the same linear combination of the actions of the group elements on the basis:

for example, if our group is C3, an element of Q[C3] looks like:

re + sa + ta2. let's call the matrix induced by a in the basis {e,a,a2}, A.

then (re + sa + ta2)(x,y,z) = (rI+sA+tA2)(x,y,z) = r(x,y,z) + s(y,z,x) + t(z,x,y) = (rx + sy + tz, ry + sz + tx, rz + sx + ty)