Why has the adjoint representation a higher dimension than the basis matrices it acts on? for example here

Why is e_1 two dim and ad(e_1) four dim?

Isn't ad(X) Y a simple matrix multiplication here? But then multiplying 4x4 with 2x2 matrices, what does it mean?

thanks
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 Think carefully. The elements of a Lie algebra are vectors. The given example happens to be gl(2), which allows those vectors to be interpreted as matrices, but the dimensionality of gl(2) is actually 4.
 Hahh! thanks genneth

Wait, still don't get it.

what the heck is, for example ad(e_1) x e_2?
How can we multiply 4 x 4 with 2 x 2 matrices?

And ae_1 + be_2 + ce_3 + de_4 is a 2x2 matrix right?
 You need to hold two views of the algebra in your head -- and it's only for this example. gl(n) are nxn matrices. But they also form a vector space, of dimension n^2. Linear operators on that vector space are NOT matrices of nxn -- they would be (n^2)x(n^2). Thus, the adjoint representation are matrices of (n^2)x(n^2), acting on a n^2 dimensional vector space.
 Recognitions: Homework Help Science Advisor This is the problem when people think too much in terms of bases. You are not mutlitpying a 4x4 matrix by a 2x2. You are letting a lie algebra acts as endomorphisms on a vector space. Think of reps, as as maps from gxV to V, in this case we send the pair (g,v) to [gv]. Things in the left slot are elements of g, and things in the right slot are elements of some vector space, which just happens to be g again in this case.
 thanks again but, how do I compute the adjoint representation in some basis (like basis given in example)? Why is my vec space n^2?
 Like I said, the elements of gl(n) are nxn matrices, which form a vector space. How many independent vectors can you find in it?

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 how do I compute the adjoint representation in some basis (like basis given in example)?

Since you didn't state what your basis was, we can't tell you. But the very definition of the adjoint rep is that

Now, what is there left to discuss?
 Here they give a basis for gl(2) from which they compute an adjoint representation. I can't see how they do it, how they go from 2x2 to 4x4 matrices.
 You seem very fixated on the 2x2 matrices. Stop it! They stop being matrices as soon as you start thinking of them as a vector space, and then they're a 4 dimensional vector space. Imagine unfolding a matrix, like this: (a b) ___> (a b c d) (c d) You can pick *any* 4 independent vectors for the basis, but the usual (1000), (0100), (0010) and (0001) are pretty good ones, as they're trivially independent, by inspection. If you run that map above backwards, you'll find them corresponding to the basis matrices given. Now, you're given the action of the adjoint representation in the original matrix view of things; construct the actual adjoint 4x4 representation by inspection on what the basis do to each other.
 Recognitions: Homework Help Science Advisor Even more explicitly, what is [e_1,e_i] for i=1..4? Now you know what ad(e_1) does to the 4 basis vectors, so you can write down the matrix for ad(e_1). It would help if you remembere to think of ad(e_1) as an endomorphism of a 4-d space.
 many thanks again, got it now by the way, I just learnt here about structure constants, why has no one brought them up yet?

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 Quote by Ratzinger .... structure constants, why has no one brought them up yet?
Because they are unnecessary for the discussion.

Why are they unnecessary? Don't they give you the matrix of adjoint representation once you have a basis?

I just found out I'm not alone being confused. Not long ago there was this thread on PF.

Especially here, that's exactly my problem (post 16)!!!

 My problem I think arises because I am trying to translate what you are saying to an explicit representation in terms of matrices (I don't know when to correctly use the word "representation" anymore) . I do see what you are saying: there is a vector space which we take as being the Lie algebra itself. A rep is a map from L x V to V. Since [x,y] does send an element of V to V, we can take this as being a possible rep, and it's called the adjoint rep. This makes complete sense to me as long as I leave it formal. Now, my problem is if I try to translate this to the language of explicit matrices. They say that one must use the structure constants to build the explicit adjoint representation. These wil represent the map LxV -> V. For sl_2(C) you gave for example [eh]=2e and so on. All I am saying is that to me it seems that the explicit adjoint representation of that algebra will be 3x3 matrices. And now, to reproduce the different commutation relations of the group, we will have to represent the elements e,f, h by the column vectors (1,0,0), (0,1,0), (0,0,1). Then the map (x,y) = [x,y] wil be reproduced in that explicit representation by simply multiplying the 3x3 matrix representing x by the column vector associated to the element y.
and what Matt Grime writes in post 18 drives it home for me!

 Let's do it (and I'll get it right this time - never do commutator relations in your head) [he]=2e [hf]=-2f [hh]=0 So ad(h) sends e to 2e, f to -2f and h to 0, so appealling to our first linear algebra course, the matrix of ad(h) is just diag(2,-2,0) in this representation (ordering the basis as e,f,h).
Got it! Finally!

(concrete examples is how most people learn, I'm no exception here...wish that view would be shared by at least one author of those trillion books on lie groups and lie algebras out there...)

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