# Questions about a projection operator in the representation theoy of groups

1. Feb 6, 2010

### bobydbcn

$$D(g)$$ is a representaiton of a group denoted by $$g$$. The representaion is recucible if it has an invariant subspace, which means that the action of any $$D(g)$$ on any vector in the subspace is still in the subspace. In terms of a projection operator $$P$$ onto the subspace this condition can be written as
$$PD(g)P=D(g)P~~~~~\forall~g\inG$$.
And furthermore the conditon can be converted into
$$D(g)P=P~~~~~\forall~g$$.
I don't know why the complex condition can be converted into the short and simple one. Can you tell me, thanks a lot.

2. Feb 12, 2010

### torquil

PD(g)P = P

which means that D(g) acts as the identity on the subspace defined by P...

I agree with your first equation. It says that application of D(g) on an element in the subspace doesn't take you outside the subspace.

But the second one looks fishy to me. It says that D(g) acts as the identity on the subspace. This is not necessary for D(g) to be an irreducible representation that acts nontrivially on the subspace defined by P.

Or am I just too tired to think straight? That may well be the case, since it's 04 o'clock right now.

Torquil

3. Feb 13, 2010

### Fredrik

Staff Emeritus
Torquil is right. The "short and simple one" can't be right. Since Pv=v for all vectors in the subspace, you'd get v=Pv=D(g)Pv=D(g)v for all v in the subspace. Perhaps you meant to write PD(g)=D(g)?

Plus, don't call D(g) a representation. D is the representation, D(g) is a linear operator. This is like calling f(x) a function, when in fact f is the function and f(x) is a number.

4. Mar 2, 2010

### bobydbcn

Now I get the answer.As we know, the condition $$PD(g)P=D(g)P$$ is a general one. But we deduce from the special one which is $$D(g)P=P$$ to the general one. Let me do it.
Becuase $$P$$ is a projection operator, we can get $$PP=P$$.
Subtituting $$D(g)P=P$$ into the above equation, we get
$$PD(g)P=D(g)P$$.
So we get the conclusion that $$D(g)P=P$$ is also a condtion, but rather abstract. Thanks a lot!

5. Mar 2, 2010

### bobydbcn

Thanks a lot! I get the right answer. Confer to my above reply.

6. Mar 2, 2010

### Fredrik

Staff Emeritus
D(g)P=P is true if and only if D(g)v=v for all v in the subspace. But PD(g)P=D(g)P is always true. So the derivation is pretty meaningless, especially considering that PD(g)P=D(g)P is very easy to prove even without that other equation.

Where did you find these equations?

7. Mar 2, 2010

### bobydbcn

I find it from the best book about lie algebra, which is called "Lie Algebras in Particle Physics, second edition", page 5. I advise you to have a look at it. Maybe it is easy for you. From your reply, I can deduce that you are really knowlegeable. Thank you.

8. Mar 4, 2010

### Fredrik

Staff Emeritus
I had a look at the book. He's saying that D(g)P=P for all g in G, for a specific G, a specific D and a specific P (G=Z3, D is given by 1.5 and P is given by 1.12), and he also says that the restriction of D(g) to the invariant subspace is the identity, which is what Torquil and I have been saying.

9. Mar 7, 2010

### bobydbcn

Thank you very much! I will go to study lie algebra. This friday, I had seen Leggett who 2003 nobel prize. I am so happy.