Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a trouble understanding an argument which I think has a simple answer but I am not getting it.

The question is that if you don't impose local fermionic symmetry the GS action has only one term and one of the equation of motion looks as $Γ\cdot p\partial_0\psi=0$, where $\psi$ is a Majorana-Weyl spinor in 10 dimensions and $Γ$ are just generalized gamma matrices.

Now $(Γ\cdot p)^2 = p^2= 0$ for the on-shell condition, but now they say that since the square is zero hence half of the eigenvalues of $Γ\cdot p$ is zero and this is where I am not getting it. If the square of a matrix is zero then how can we say that half of it's eigenvalues are zero. Since the eigenvalues of a squared matrix is square of the eigenvalues of the matrix whose square is taken, so how can we say that half of the eigenvalues are zero? May be it can happen when spinor contain complex component but here the spinor is Majorana too ( thinking in some different way that how it will impose constraint on half of the component of $\psi$ but this is not working either).

(For reference, see chapter 5 of GSW Vol.1.)

share|cite|improve this question

If a linear map $A:V\to V$ in a finite vector space $V$ squares to zero

$$\tag{1} A^2~=~0,$$

then the image of $A$ must lie in the kernel of $A$:

$$\tag{2} {\rm Im}(A)~\subseteq~{\rm Ker}(A),$$

and therefore

$$\tag{3} \dim {\rm Im}(A) ~\leq ~\dim {\rm Ker}(A).$$

Now we also know that

$$\tag{4} \dim {\rm Im}(A) ~= ~\dim V - \dim {\rm Ker}(A). $$

Together (3) and (4) yield that the dimension of the kernel of $A$ is at least half the dimension of the full space $V$,

$$\tag{5} \dim {\rm Ker}(A)~\geq ~ \frac{1}{2}\dim V , $$

which implies that at least half the eigenvalues of $A$ are zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.