# Diagonalize mass matrix term for fermions and “doubling trick” in m(atrix) theory

Can someone help me understand the "Doubling trick" at page 36 in http://inspirehep.net/record/887513/files/sis-2002-060.pdf (named "Scattering in Supersymmetric M(atrix) Models" by Robert Helling) or help me in some other way to get the mass for the fermions from the given Lagrangian, preferably without knowing the explicit form of the SO(9) gamma matrices?

• Rather than link against the thesis, it would be better to put the trick/equation inlined here (with a reference to the text). Also, what is your confusion about it? Where it comes from, how to use it, why it works, etc? – tpg2114 Nov 2 '13 at 17:54

Let $M$ be the mass matrix for fermions $\psi_+$ and for $\psi_-$ (separately). It is obtained by $\not{D}\not{D^+}= -\partial_t^2+ M^2$

Then $M^2=r^2 \mathbb{Id_{16}}- \not{v}$, Now, the $16*16$ matrix $\not{v}$ has a zero trace, and it square is $\vec v^2 Id_{16}$, so the only possibility is that the matrix $\not{v}$ has 8 eigenvalues $v$, and 8 eigenvalues $-v$ (here $v$ means $\sqrt{\vec v^2}$). So the matrix $M^2$ has 8 eigenvalues $r^2+v$ and 8 eigenvalues $r^2-v$. This is true for $\psi_+$ and for $\psi_-$, while $\psi_3$ is obviously massless.

[EDIT]

The gamma matrices of $SO(9)$ are real, so $\not{B}$ is hermitian. $\partial_t$ is antihermitian (because $i\partial_t$ is hermitian), so starting with $\not{D}=\partial_t-\not{B}$, it is easy to see that $\not{D}^{\dagger}=-\partial_t-\not{B}$

If you neglect order 3 terms in the Lagrangian ($\psi^2Y, \psi^2A$), and apply Lagrange equation on $\psi_+$, you get $\not{D}\psi_-=0$. And, because $\psi_+ = (\psi_-)^*$, and $\not{B}$ is real, you have also $\not{D}\psi_+=0$

The mass matrix apply separately to $\psi_+$ and $\psi_-$, simply because $\psi_+ = (\psi_-)^*$, and the mass matrix is real.

• Why do we have a mass matrix for both $\psi_+$ and $\psi_-$ separately? And how come that if $\not{D}=\partial_t-\not{B}$ then $\not{D}^{\dagger}=-\partial_t-\not{B}$? And where do $\log$ come from? If it's from the effective action $\Gamma^{(1)}=\frac{1}{2}\text{Tr}\log (\mathscr{O})$ does that mean that $\not{D}$ is the wave operator for the fermions, and how do I see this? – Natanael Nov 6 '13 at 8:26