Is anyone familiar with this algebra from here (at 12:00), which seems to be related to Lie decomposition of the SU(5) group?
$$ 5 \bigotimes 5 = 10 \bigoplus 15$$
Other examples given are:
$$ 5 \bigotimes \bar 5 = 1 \bigoplus 24$$ $$ \bar 5 \bigotimes 10 = 5 \bigoplus 45$$ $$ 5 \bigotimes 10 = 40 \bigoplus \bar {10}$$ $$ 10 \bigotimes 10 = 50 \bigoplus 45 \bigoplus 5$$
Could anyone explain the step-by-step algebra in order to get at this equality $ 5 \bigotimes 5 = 10 \bigoplus 15$, please?
Let $V=\Bbb C^5$ be the "natural" representation of $SU(5)$, with $\bar V$ the conjugate representation. Then
$V\otimes V$ splits into $V\odot V$ and $V\wedge V$, symmetric and skew-symmetric products of dimensions 15 and 10. There was probably a print error, typo.
$V\otimes \bar V$ splits in range and kernel of $a\times b \mapsto b^Ta$
The 10 representation is, as per above, $V\wedge V$. Then the split-defining operation on $\bar V\otimes (V\wedge V)$ is $a\otimes (b\wedge c)\mapsto (a^Tb)c-(a^Tc)b$, with the range being all of $V$.
$V\otimes(V∧V)$ splits into kernel and range of $a\otimes (b\wedge c)\mapsto a\wedge b\wedge c$, where $\Lambda^3V$ has dimension $10$.
$(V∧V)\otimes(V∧V)$ similarly splits into kernel and range of $(a\wedge b)\otimes (c\wedge d)\mapsto (a\wedge b)\wedge (c\wedge d)$, where $\Lambda^4V$ has dimension $5$.