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Fearless Symmetry by Robert Gross, Avner Ash

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THE GREEKS HAD A NAME FOR IT 171
We will leave the rest of the computations to you.
EXERCISE: Show that R is a faithful representation.
What is the character of the representation R? Here we must
remember that χ
R
will be an element of F
2
, not of Z.Soweget
χ
R
(σ ) = 1, χ
R
(τ ) = 0, and (here is the surprise) χ
R
(e) = 1 + 1 = 0.
Again, we can check that χ
R
is constant for all members of
a conjugacy class: We see that χ
R
(τ ) = χ
R
(τ σ ) = χ
R
(τ σ
2
), and
χ
R
(σ ) = χ
R
(σ
2
).
How the Character of a Representation Determines the
Representation
The amazing thing is that the character χ
r
of a representation r
determines r in an appropriate sense, so it is in fact a distinct token
of the representation. The character is sometimes easier to think
about than r itself, because it assigns a single number to each
element of the source group, rather than a whole matrix.
To see in what sense a character determines a representation,
perform the following experiment. Write down any 3-by-3 matrix A
and find its trace. Now pick your favorite invertible 3-by-3 matrix B
and compute BAB
1
. You will see that the trace of the matrix BAB
1
is the same as the trace of A! Here is an example, but you should do
your own. We will let
B =
31065
41491
01 7
,
and then we can compute that
B
1
=
7 50
28 21 13
4 32
.
172 CHAPTER 15
You can check that BB
1
= B
1
B = I.WepickanA at random:
A =
235
71113
17 19 23
.
It is easy to see that the trace of A is 2 + 11 + 23 = 36. Finally, we
have
BAB
1
=
31065
41491
01 7
235
71113
17 19 23
7 50
28 21 13
4 32
=
23, 085 17, 609 14, 322
32, 309 24, 645 20, 045
2, 454 1, 872 1, 524
,
and sure enough the trace of BAB
1
is 23, 085 + 24, 645 1, 524 =
36. This is the reason that elements in a single conjugacy class have
the same character as one another, whatever representation r you
choose. Namely, if x and y are in the same conjugacy class, then
for some z we have the equation z x z
1
= y, so that r(z) r(x)
r(z)
1
= r(y), so r(x) and r(y) have the same trace.
So if r is our representation, then Br(g)B
1
will have the same
trace as r(g).
2
We now ask the question: To what extent can we
reconstruct the representation if all we know is its character? It
looks as if the character of r could hardly determine r. Just change
r by picking an invertible matrix B
g
for each g, replace r(g)by
B
g
r(g)B
1
g
, and the traces, and hence the character, will stay the
same.
But wait a minute. Why should the function g B
g
r(g)B
1
g
be a
morphism? In general, it will not be.
However, if we choose B
g
to be a constant invertible matrix B,
and define s(g) = Br(g)B
1
for every g in H, then you can check that
s is itself a morphism.
EXERCISE: If r(g) is a linear representation of a group H (so
that r(g h) = r(g)r(h)), and B is a constant invertible matrix
2
Note that Br(g)B
1
is the product of three matrices: B, r(g), and B
1
.
THE GREEKS HAD A NAME FOR IT 173
of the same size with entries from the same number system,
and s(g) = Br(g)B
1
, show that s(g h) = s(g)s(h).
SOLUTION: We write it out:
s(g)s(h) = (Br(g)B
1
)(Br(h)B
1
) = (Br(g))(B
1
B)(r(h)B
1
)
= Br(g)r(h)B
1
because BB
1
= I. And the last term equals Br(g h)B
1
because r is a morphism, and in turn Br(g h)B
1
= s(g h).
In summary, s(g h) = s(g)s(h) for any g and h in H,sos is a
morphism—in fact, it is another linear representation of H.
Now, the representation s is essentially the same as r—it is
just carrying around a B in front and a B
1
in back. If you
know linear algebra, then you know that s represents the same
geometry as r, except with respect to a different basis. If you
change your coordinate axes, you will change the numbers that
represent events, but you do not change anything in the reality
itself. Similarly, if you replace r by s, you are making a superficial
change in your description, but the underlying representation
is essentially the same. Of course, it is not exactly the same,
so we need a new word. We say that r and s are equivalent
representations.
Take our example of r from above. Let s(g) = Br(g)B
1
, where B
is the matrix we used above. Then we can compute that
s(σ ) = Br(σ )B
1
=
31065
41491
01 7
010
001
100
7 50
28 21 13
4 32
=
411 292 19
581 413 24
53 38 2

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