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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