# Understanding conclusion of Proof that a matrix that commutes with all gamma matrices is proportional to the identity matrix

I am trying to understand why a matrix $$M$$ that commutes with all gamma matrices $$\gamma^\mu$$ is proportional to the Identity matrix. I am following example 3.18 in Voja Radovanovic's book on solved QFT problems. considering the set of 16 linearly independent Gamma matrices: $$\Gamma^a={\{I,\gamma^\mu,\sigma^{\mu\nu},\gamma_5,\gamma_5\gamma^\mu\}}$$ I understand most of the procedure. I begin with the fact that I can write any Matrix as a sum of the Gamma matrices and separate one of them. (with $$\Gamma^b\neq I$$): $$M=\sum_{a}c_a\Gamma^a=c_b\Gamma^b+\sum_{a\neq b}c_a\Gamma^a$$ Now I multiply from the left by $$\Gamma_d$$ and from the right by $$\Gamma^d$$, such that $$\Gamma^d$$ and $$\Gamma^b$$ anticommute: $$\Gamma_d M\Gamma^d=c_b\Gamma_d\Gamma^b\Gamma^d+\sum_{a\neq b}c_a\Gamma_d\Gamma^a\Gamma^d$$

Since $$M$$ by definition commutes with all $$\gamma^\mu$$, and all $$\Gamma^d$$ are comprised of said $$\gamma^\mu$$, $$M$$ commutes with any $$\Gamma^d$$. Also, $$\Gamma^b$$ will anticommute with $$\Gamma^d$$, and $$\Gamma_d\Gamma^d=1$$: $$M=-c_b\Gamma^b+\sum_{a\neq b}c_a\Gamma_d\Gamma^a\Gamma^d$$ But I know that the product of any two $$\Gamma$$ simply give another $$\Gamma$$ up to a factor $$\eta=+-1$$ or $$\eta=+-i$$ $$M=-c_b\Gamma^b+\sum_{a\neq b}c_a\eta\Gamma^a$$ Here, I can multiply the first equation as well as the previous one by $$\Gamma_b$$ and take the trace, giving the following two expressions: $$Tr(M\Gamma_b)=Tr(c_b\Gamma^b\Gamma_b)+Tr(\sum_{a\neq b}c_a\Gamma^a\Gamma_b)$$ $$Tr(M\Gamma_b)=Tr(-c_b\Gamma^b\Gamma_b)+Tr(\sum_{a\neq b}c_a\eta\Gamma^a\Gamma_b)$$ Clearly, $$c_b$$ has to be 0 for the first two expressions to be equivalent. But the book now says that it is "obvious" that all $$c_a$$ terms are $$0$$ except for the one multiplying the identity matrix, but I do not see how I could draw that conclusion. How can we know that?

• Consider to define what you mean with $\Gamma$. Jan 5 at 18:36
• By $\Gamma$ I mean one of the 16 dirac gamma matrices. I will edit it in Jan 5 at 18:42
• Are you familiar with Shur's lemma? en.wikipedia.org/wiki/Schur%27s_lemma Jan 5 at 18:47
• "I begin with the fact that I can write any Matrix as a sum of the Gamma matrices" This is not true. You can write any 4x4 matrix as a sum of the Gamma matrices. It is a convenient fact that the 16 Gamma matrices turn out to be independent in this sense.
– hft
Jan 5 at 18:50
• I am not, but I will read it now. How does it relate to my question? Jan 5 at 18:50