# Show that $v_iw_j + v_j w_i$ is a representation of $SU(2)$

I understand that the mathematical (i.e. not how my physics professor sees it) definition of a representation is simply a homomorphic mapping from the group $G$ to GL$(n, V)$. However, my professor does not use the term representation this way. For example, he defines the fundamental representation of SU(2) as

The fundamental representation: we define the 2-dimensional representation as a column vector of 2 complex elements which transforms under the algebra, under usual multiplication. We call this $v_i$ where $i=1,2$. The algebra acts on the representation. Under this action, $$v_i \rightarrow v_i + iM_{ij}v_j,$$

where $M_{ij}$ is a matrix in the algebra.

My understanding:

(1) I assume here that he means "representation space", and not "representation", but I'm not sure if this assumption is exactly right.

(2) By transforming "under the algebra", does he mean "transforms under infinitesimal SU(2) transformations"? If so, is that exactly right? Since would a transformation under the algebra look more like $v_i \rightarrow iM_{ij}v_j$ ?

(3) If (1) is the case, then why on earth would somebody even care about the representation space to begin with? It alone doesn't tell us anything about the group.

Main Question:

Show that the symmetrize combination $v_iw_j + v_jw_i$ is a representation. i.e. the transformation moves the fields into linear combinations within the representation.

My attempt

First, I assume that he means representation of SU(2) (?). Anyways, I am able to show that the product transforms like

$$v_iw_j \rightarrow v_iw_j + iw_jM_{ik}v_k + iv_i M_{jk}w_k .$$

Applying this to $v_j w_i$ and adding, I am able to show that

$$v_i w_j + v_j w_i \rightarrow v_iw_j + v_j w_i + iM_{ik}(v_jw_k + w_jv_k) + iM_{jk}(v_iw_k + w_iv_k).$$

I know that this is also symmetrized, but that is about all I know. I have no idea how this is supposed to tell me whether or not this is a representation (or representation space (?), referring back to (1), (3)). The fact that I don't know where to go with this, leads me to believe that I must have some fundamental misunderstanding in the form of the following question:

How would I show that $v_i w_j + w_i v_j$ is a representation, given the definition in the first block quote?

I like this question. First of all, welcome to the sloppy mathematical language of (some if not most) physicists. It is a nightmare for me, as I am really fond of mathematics. I will provide you with a translation from this language into the language of mathematics:

we define the 2-dimensional representation as a column vector of 2 complex elements which transforms under the algebra, under usual multiplication.

DEFINITION: We define the two-dimensional fundamental representation of the Lie group $\text{SU}(2,\mathbb{C})$ on the vector space $\mathbb C^2$ as a homomorphic (isomorphic) mapping $\rho$ from the group $\text{SU}(2,\mathbb{C})$ to itself.

By acting with the representation morphism on the representation space $\mathbb{C}^2$, we define the $\text{SU}(2,\mathbb{C})$ fundamental spinor vector $\psi$ (rather its two components with respect to the canonical basis in $\mathbb{C}^2$) as

$$\text{SU}(2,\mathbb{C})\ni U\mapsto U \in \text{SU}(2,\mathbb C), \forall \psi_i\in\mathbb C^2, (\psi_U)_i = \sum_j U_{ij} \psi_j$$

By "usual multiplication" he means that the composition of homomorphisms is actually reduced to 2x2 matrix multiplication. From what you quoted, there is no reason to consider the Lie algebra $\mathfrak{su}(2,\mathbb C)$, thus no "infinitesimal operations" anywhere.

For the problem at hand, you need to transpose the mathematical abstract definitions into 2x2 and 4x4 matrix algebra. What you are asked to prove is that the symmetrized tensor/outer product of two fundamental spinors is a genuine tensor. This means that:

Write $v_i w_j = \frac{1}{2} (v_i w_j + v_j w_i) +\frac{1}{2} (v_i w_j - v_j w_i)$. Then the antisymmetric part is the trace times the SU(2)-invariant antisymmetric tensor $\epsilon_{ij}$ thus under the group is an invariant. The symmetric part - I invite you to write it as a combination of a product of v,w and the 3 Pauli matrices.

• Does the fact that he is saying the transformation is $v_i \rightarrow v_i + iM_{ij}v_j$ not imply that the transformation is your $U_{ij}$ expanded around identity? Then we would be considering the algebra, no? – InertialObserver Nov 10 '17 at 0:49
• Thinking about it, yes. But there was no mention of the Lie product (commutator), so I just assumed it was only about group elements. – DanielC Nov 10 '17 at 0:51