# SU(2) index convention

1. Feb 7, 2009

### intervoxel

I'm studying Lie groups using G. 't Hooft's tutorial:

http://www.phys.uu.nl/~thooft/lectures/lieg.html

In page 34, he introduces superscript and subscript indices starting from equation 6.27:

$$\phi_{\alpha} \rightarrow \phi_{\alpha}' = X_{\beta}^{\alpha}\phi^{\beta}$$

What is the meaning of the X operator receiving both indices?

P.S.: Please take into consideration that I am a self teaching student :)

2. Feb 7, 2009

### HallsofIvy

Staff Emeritus
The usual convention (the "Einstein summation convention) is that when the same index appears in both superscript and subscript it indicates a sum over that index. If you think of $\phi^\beta$ as indicating the components of a column matrix, then $X_\beta^\alpha[itex] would be a matrix with [itex]\alpha$ enumerating the columns and $\beta$ the rows. $X_\beta^\alpha\phi^\beta$ then would be the matrix multiplication of the two matrices, giving a row matrix as indicated by the $\alpha$ subscript on $\phi_\alpha$.

Notice I said "think of". These are functions NOT matrices but you are just generalizing matrix concepts.

3. Feb 7, 2009

### intervoxel

I'm still a bit confused. On that same page, he says the only allowed Kronecker delta function is $$\delta_{\beta}^{\alpha}$$ (and not $$\delta^{\alpha\beta}$$).

How should I interpret this statement?

4. Feb 7, 2009

### xepma

The statement he makes is about invariant tensors. These are tensors which do not transform under the action of an element from the Lie group.

In this case the Lie group is SU(2). The elements act on the representation space through the matrices X. Suppose we start out with the Kronecker delta with two upper indices, $\delta^{\alpha\beta}$. If such a tensor is an invariant tensor with respect to the Lie group, then the Kronecker delta retains it's properties under the action of the Lie group. In this case we would want:

$\delta^{\alpha'\beta'} = X^{\alpha'}_{~\alpha}X^{\beta'}_{~\beta}\delta^{\alpha\beta}$

But this statement is correct only if

$\sum_{\alpha}X^{\alpha'}_{~\alpha}X^{\beta'}_{~\alpha} = \delta^{\alpha'\beta'}$

which is the same as stating:

$(X^{-1})^{\alpha}_{~\beta} = X_{\beta}^{~\alpha}$ (note the subtle change in the order of upper and lower indices).

This last statement might be a bit puzzling so try to reproduce it.

In any case, this property must hold for all X, i.e. for all X(g), where g is an element of the Lie group. But for SU(2) we already know that we are dealing with unitary tensor. For unitary tensors the relation between X and X^{-1} is a different one then the one mentioned above (do you know which one).

So what does this mean? It means that the Kronecker delta with two upper indices is not invariant under the action of the Lie group SU(2). If we start out with this tensor then its form changes when we perform a SU(2) transformation, and we are left with a completely different tensor, i.e. one that doesn't have the properties of the Kronecker delta. In this perspective, the Kronecker tensor is not treated on some special footing with respect to SU(2). In particular, it is not an inveriant tensor.

If you thing you understand it, try to see if you can prove that the tensors $\delta^{\alpha}_{\beta}$ and $\epsilon^{\alpha\beta}$ are indeed invariant tensors with respect to SU(2).

Hint, repeat the steps I performed for $\delta^{\alpha\beta}$ but now for these two cases, and try to come up with a restriction on X. What is the restriction, and is it satisfied by all elements of SU(2)?

5. Feb 7, 2009

### jambaugh

There are two dual conjugate representations of the SU(N) groups and their SL(N;C) complexification, the vector and dual vector representations. We represent components of one with raised indices and of the other with lowered indices. You can contract (sum the product of dual components) to get a scalar which invariant under group transformations on the vector and dual vector.

In term of components you get $\psi^a \phi_a = s$ (summation over $a$ assumed). If you act with a group element:
$$\psi \to \psi',\quad \phi \to \phi'$$
the group acts dually on these so that:
$$\psi^a \phi_a = s={\psi'}^a{\phi'}_a$$

We can define a dual vector as a linear functional mapping vectors to scalars and contrawise vectors map dual vectors to scalars...two ways of saying a vector and dual vector contract to a scalar.

Now higher rank objects (tensors) are elements of product spaces of these two spaces and their component thereby have multiple indices so keeping the upper and lower indices straight is how we keep straight how the quantities transform under the group and making sure contracted quantities are not basis dependent.

In the group SL(N;C) there is no prefered metric so getting a number from two vectors by say:
$n = \sum_a \psi^a \phi^a$
1.) Implies some inner product where-in the basis in which these components were chosen is ortho-normal. In short it is picking a metric defined by this basis and is not basis independent.
2.) Results in the quantity n not being unchanged when we act on the two vectors with a particular group element.

Contrawise since a vector and dual vector by definition transform dually the quantity s above will be the same even if you transform the psi and phi vector, dual vector pairs.

So we make it a rule never to contract (sum over) two upper or two lower indices but only one upper with one lower.

The Kronecker delta is basically the components of the identity operator:
$$\delta^a_b \psi^b = \psi^a$$
$$\delta^a_b \phi_a = \phi_b$$
We want it to be the identity operator no matter what basis and to map vectors to vectors and dual vectors to dual vectors. So one index needs to be up and the other down.

If we always stick to this notational convention then we are assured that any unindexed quantity is a true scalar= invariant under group transformations. This is good practice whether we are working in a unitary group of a quantum theory or the Lorentz group for space-time vectors in classical relativity or general abstract mathematics of linear representations of any group.

You may find the example from calculus helpful. Given coordinates of a position vector $$x^\mu$$ the partial derivatives are components of a dual vector (the gradient)
$$\partial_\mu = \frac{\partial}{\partial x^\mu}$$
Observe how each transform under a change of coordinates say
$$x^1 \mapsto y^1 = 3x^1$$
$$\frac{\partial}{\partial x^1} \mapsto \frac{\partial}{\partial y^1}= \frac{1}{3} \frac{\partial}{\partial x^1}$$