What is the $10$ in the $\mathbf{4}\otimes\mathbf{4}$ tensor product of $SO(6)$? This is the question 22.D in Howard Georgi's Lie Algebras book, I thought about for a minute, but couldn't come up with a plausible answer.
It's a fact that the SO(6) and SU(4) algebras are equivalent. Further, the fundamental $\mathbf{4}$ of SU(4) is the spinor of SO(6). The relevant part of the question is "In SU(4), $\mathbf{4}\otimes\mathbf{4}=\mathbf{6}\oplus\mathbf{10}$, the $\mathbf{6}$ is the vector representation of SO(6). What is the $\mathbf{10}$ in the SO(6) language?"
The fact that $\mathbf{4}\otimes\mathbf{4}$ must contain the $\mathbf{6}$ is obvious since two spinors can be combined to give a vector by definition. I am a bit puzzled with the $\mathbf{10}$ though. In highest weight notation of the SO(6), where the $\mathbf{6}=[1,0,0]$, the spinor is $\mathbf{4}=[0,1,0]$, and we can show that the $\mathbf{10}=[0,2,0]$. But what is its interpretation?
 A: In the $SU(4)$ language, the 10-dimensional representation is the symmetric spintensor $T_{(ab)}$ with $4\times 5 / (2\times 1) = 10$ components.
In the $SO(6)$ representation, it is the self-dual 3-form with 
$$ \frac 12 \cdot \frac{ 6\times 5 \times 4}{3\times 2 \times 1}  = 10$$
components. It's the tensor $T_{[kmn]}$ that also obeys
$$ T_{kmn} = \frac{\pm i}{3!} \epsilon_{kmnpqr} T_{pqr} $$
I think that in the Euclidean signature, the factor of $\pm i$ (the sign is self-duality vs anti-selfduality) has to be added because the doubly repeated dualization is $*^2=-1$. Note that due to the $i$, the 10-dimensional representation is complex, not real (and not pseudoreal).
You may guess the rep because it's the smallest irreducible representation of $SO(6)$ larger than the 6-dimensional one.
As user23... pointed out, you may also derive this fact by sandwiching gamma matrices. All the components of a tensor product of the spinors may be obtained as $\psi^T M \psi$ for some matrix $M$ inserted between the two spinors (those we tensor multiplied). Every $M$ may be written as a combination of products of gamma matrices. Because none of the $\psi$ in $\psi^T M\psi$ is complex conjugated, I need an odd number of gamma matrices (multiplied and antisymmetrized), and 1,3,5 are the only options. 1,5 are equivalent and 3 is the only choice (equivalent to itself, which is why the self-duality is possible).
Note that the 20-dimensional representation $T_{kmn}$ was divided to two pieces, the complex self-dual and complex anti-selfdual representation. You might be surprised that the 6-dimensional representation (the antisymmetric tensor product of the two spinors) is real. But it is real because of another duality operation, now in the $SU(4)$ language. Even though $a,b$ are "complex" $SU(4)$ indices and $A_{[ab]}$ "looks" complex as well, the real condition on the $A_{[ab]}$ antisymmetric spintensor may be written as $A^* = *_4 A$. (An added relative sign would just make the 6-component representation "pure imaginary" which is basically the same as "real" when it comes to the search for reps.)
