Question

Group Theory

"Using without proof, the homomorphism theorem, or otherwise, show that
U(N)/SU(N) is isomorphic to  U(1)."

I really struggle with group theory, so if anyone is willing to take me through this slowly , I would be very grateful!

Answer 1

My worst nightmare as a child: Group Theory.

Answer 2

@primeralph I know the exact feeling..I just can't get my head around it at all

Answer 3

@sarahusher I could send you a pretty good book for this, but I can't help. Can barely remember how to solve this.

Answer 4

I'm unfamiliar with the notation U(N), and SU(N). What are these called/defined as?

Answer 5

what does '"without proof" mean?

Answer 6

whats the name of the book i wanna learn this stuff

Answer 7

if i am not mistaken (which i may very well be) an element of \(U(1)\) is a complex number \(e^{i\theta}\)

Answer 8

@dan815 Hold on, it's somewhere in a hard drive; I had it with Lie Theory for a while, back in High School.

Answer 9

@sarahusher Any idea what the notations mean?

Answer 10

What part don't you understand - do you need help with the definitions of U(n) and SU(n)?

Answer 11

hello?

Answer 12

U(N) in defined as the Unitary Matrix such that
\[U(N)=\left\{ A \in M _{N}(Complex): A ^{t}=A ^{-1} \right\}\]

and SU(N) is defined as the Special Unitary Matrix such that:
\[SU(N)=\left\{ A \in U(N): detA=1 \right\}\]

Answer 13

and yes @primeralph that book would be great!

Answer 14

Basically, you want to show the existence of a homomorphism \(\varphi:U(N)\to U(1)\) such that \(\ker(\varphi)=SU(N)\). I think the homomorphism you want to use is \[\begin{aligned}\varphi:U(N)&\to U(1)\\A&\to\det(A)\end{aligned}\]

Answer 15

again maybe i am confused
but the obvious map would send an element of \(U(n)\) to the determinant, and kernel would be the matrices with determinant 1

Answer 16

better expressed by @KingGeorge

Answer 17

Consider the determinant map $$\det:\mathrm{U}(n)\to \mathrm{U}(1)$$ The kernel of which is (by definition) SU(n). Apply the first homomorphism theorem.

Answer 18

Hi @Jack172
Sorry I am here, I'm in China at the moment, and the internet is very tempremental so can't decide if it'll let me on the website or not haha

Basically I'm confused by the 'quotient' aspect of it, [ie U(N)/SU(N)] and what U(1) is?
and how you would even start this off...

Answer 19

Don't get confused by the quotient, it's just notation. In this case, you don't even have to work with it at all. Just prove that \(SU(N)\) is the kernel of the determinant homomorphism, and you'll be able to apply the fundamental theorem of homomorphisms to immediately prove what you want.

Answer 20

and think of U(1) as just the complex numbers with magnitude 1.

Answer 21

in reality \(G/M\) is a set of cosets, but you can think of it as putting everything in the same bucket (equivalence class) if they have the same feature modulo \(M\)
since here \(M=SU(N)\) i.e. matrices with determinant 1, the the "elements" of \(G/M\) only differ by the determinant

okay that was a rather lousy way to put it, but it is late

Answer 22

@KingGeorge Sorry what do you mean when you say 'prove SU(N) is the kernel of the determinant homomorphism?"

[when I say I don't understand Group Theory I really don't!) sorry, if this is basic stuff I'm asking!

Answer 23

the kernel is the set of elements that are sent to the identity, and the identity in \(U(1)\) is the number 1

Answer 24

so the kernel is by definition \(SU(N)\)

Answer 25

Alright. So the kernel of a group homomorphism \(\varphi:G\to H\) is defined as \(\ker(\varphi):=\{g\in G\,|\,\varphi(g)=1\in H\}\). In this case, the identity of \(U(1)\) is equal to \(\left[1\right]\). Or just 1.

Answer 26

you also have to say that the determinant map is a homomorphism, but that is clear by the mantra "the determinant of the product is the product of the determinants"

Answer 27

So you want to show that \(A\in SU(N)\) if and only if \(\det(A)=1\). This is basically true from the definition of \(SU(N)\).

Satellite brings up a good point. You may want to show the the determinant is indeed a homomorphism.

Answer 28

by which @KingGeorge means the map \(\phi (A)=\det(A)\) is a homo

Answer 29

When you say showing the determinant is a homomorphism, do you just mean that:

det(AB)=det(A)det(B) ?

Answer 30

yes

Answer 31

Yes. Although you can probably take that on faith.

Answer 32

in general the idea for all such problems is this
if you want to show that \(G/H\cong G'\)
find a (usually obvious) map \(\psi\) from \(G\to G'\) where everything in \(H\) gets sent to the identity of \(G'\)

Answer 33

it should be the first thing that springs to mind more or less

Answer 34

okay, also if U(1) is the complex numbers with magnitude 1
what is an example of SU(2)...

I think the notation isn't helping me, cause I can't visualise it

Answer 35

my understanding would be 2 by 2 unitary matrices with determinant 1

Answer 36

@KingGeorge seem right?

Answer 37

seems right to me.

Answer 38

right okay, I get that

Answer 39

\(U(1)\) is "one by one" unitary matrices, i.e. complex numbers of magnitude 1

Answer 40

and SU(3) would be a 3x3 unitary matrix with det=1
and so on...?

Answer 41

correct.

Answer 42

that is my understanding, yes

Answer 43

and what is \[Z _{2}\], is that 2x2 matrices with integers?

Answer 44

no, \(\mathbb{Z_2}=\{0,1\}\)

Answer 45

under addition.

Answer 46

with the operation addition mod 2

Answer 47

also known as \(\mathbb{Z}/2\mathbb{Z}\)

Answer 48

probably the simplest ever version of such a proof would be to show that
\[\mathbb{Z}/\mathbb{2Z}\cong \mathbb{Z_2}\]

Answer 49

Okay thankyou so much @satellite73 and @KingGeorge for your help [and patience] :)

I'm going to try and work through a few others, and try and get used to proving these things... I think I need to spend a bit of time looking at examples of what each notation means

Answer 50

You're welcome, and good luck.