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Suppose $\chi$ is a nonzero, non-trivial character of $G$ and $\chi(g)$ is a non-negative real number for all $g \in G$. Prove that $\chi$ is reducible.

My try : I want to prove $\langle\chi,\chi\rangle>1$ then we are done.

Now, $$\langle\chi,\chi\rangle=\frac 1{|G|}\sum_{g \in G}\chi(g)\overline{\chi(g)}=\frac 1{|G|}\sum_{g \in G}\chi(g)^2>0$$ How do I know that $\sum_{g \in G}\chi(g)^2>|G|$?

Any help?

One note : The chapter that I am reading didn't prove Schur's Orthogonality yet so I don't want to overkill the problem. I don't know whether the solution will actually involve the steps of Schur's Orthogonality relation or not!

If so please mention that then I will look into this later. I just came to know the name after seeing the other related solutions.

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I presume you are working over $\mathbb C$.

$\chi(1)$ must be positive, so if $\chi(g)$ is real and nonnegative for all $g$, then $d=\frac{1}{|G|}\sum \chi(g)$ is a positive real. Now if a representation $V$ affords $\chi$, then $d$ is exactly the $\mathbb C$-dimension of $V^G$. But $V^G$ is a $G$-submodule of $V$ on which $G$ acts trivially. Hence $V$ has a positive-dimensional submodule on which $G$ acts trivially. The conclusion is that $V$ is reducible or trivial.

If you permit yourself to use orthogonality, the argument is easier: $\langle \chi, \chi_1\rangle$ is positive, so $\chi$ is reducible or $\chi=\chi_1$ is trivial.

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  • $\begingroup$ Yes I saw your last comment before in one answer. By the way, is $V^G=Hom_{\Bbb CG}(V,V)$? $\endgroup$ – Ri-Li Nov 12 '19 at 4:05
  • $\begingroup$ 1) I still have doubts why are you taking $d=\frac{1}{|G|}\sum \chi(g)$ is a positive real where $<\chi,\chi>=\frac{1}{|G|}\sum \chi(g)^2$ ?...... 2) "Now if a representation $V$ affords $\chi$, then $d$ is exactly the $\mathbb C$-dimension of $V^G$. But $V^G$ is a $G$-submodule of $V$ on which $G$ acts trivially. Hence $V$ has a positive-dimensional submodule on which $G$ acts trivially. The conclusion is that $V$ is reducible or trivial."---------I don't know this result yet. I am reading from James and Liebeck. I am on chapter 14. $\endgroup$ – Ri-Li Nov 12 '19 at 4:16
  • $\begingroup$ If you just want to prove that $d$ is the dimension of $V^G$ (the subspace fixed points of the action), you can consider the endomorphism $p = \frac{1}{|G|}\sum g$ of $V$. It's easy to compute that $p^2=p$ so $p$ is a projection. Then try to see that the fixed points of $p$ are exactly $V^G$, and conclude using $d=Tr(p)$. $\endgroup$ – Captain Lama Nov 12 '19 at 5:24
  • $\begingroup$ Yeah, I got it later. Thanks a lot! $\endgroup$ – Ri-Li Nov 12 '19 at 6:33

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