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Well, I have a character table of group $S_4$: \begin{array}{rrrrrrrrrrr} & \chi_1 & \chi_2 & \chi_3 & \chi_4 & \chi_5 \\ e & 1 & 1 & 2 & 3 & 3 \\ (12) & 1 & -1 & 0 & -1 & 1 \\ (12)(34) & 1 & 1 & 2 & -1 & -1 \\ (123) & 1 & 1 & -1 & 0 & 0 \\ (1234) & 1 & -1 & 0 & 1 & -1 \\ \end{array} And I have a representation of $S_4$: $V=\chi_3^{\otimes 2} \otimes \chi_4^{\otimes 2} \oplus \chi_5^{\otimes 3}$. So, I need to find the character of this representation, and I've done this and I got: \begin{array}{rrrrrrrrrrr} & V \\ e & 63 \\ (12) & 1 \\ (12)(34) & 3 \\ (123) & - \\ (1234) & -1 \\ \end{array} And then I must find a character of $\Lambda^2 V$ and I've got a known formula to do it: $\chi_{\Lambda^2 V}(g)=\frac{\chi_{V}(g)^2-\chi_{V}(g^2)}2$ then I got: \begin{array}{rrrrrrrrrrr} & \Lambda^2 V \\ e & 3906 \\ (12) & -31 \\ (123) & 0 \\ (12)(34) & 4 \\ (1234) & -2 \\ \end{array} And I know that the character should be a integer number, but I got $-\frac{3}2$. Can someone explain where my mistake? May be I wrong counted a character of $V$.

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  • $\begingroup$ What actually is $W$? To have $\frac12(\chi(e)^2-\chi(e^2))=0$ you need $\chi(e)=0$ or $1$; the representation is zero or one-dimensional. $\endgroup$ – Lord Shark the Unknown Dec 9 '17 at 14:13
  • $\begingroup$ @LordSharktheUnknown I've a mistake, $V=W$, I'm corrected this. $\endgroup$ – Mathworld Dec 9 '17 at 14:14
  • $\begingroup$ That's a peculiar character table. The group $S_4$ has irreps of degrees $2$ and $3$, not just $1$. $\endgroup$ – Lord Shark the Unknown Dec 9 '17 at 14:16
  • $\begingroup$ I thought that $0-0=0$, not $-1$. $\endgroup$ – Lord Shark the Unknown Dec 9 '17 at 14:17
  • $\begingroup$ @LordSharktheUnknown My bad, it is one more mistake, I've corrected. $\endgroup$ – Mathworld Dec 9 '17 at 14:17

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