# Tag Info

9

Among normal books, Becker-Becker-Schwarz probably matches your summary most closely. However, you may want to look at a list of string theory books: http://motls.blogspot.com/2006/11/string-theory-textbooks.html Don't miss the "resource letter" linked at the bottom which is good for more specialized issues such as string field theory. An OK review of ...

6

The BFSS matrix model is a quantum mechanical model – i.e. quantum field theory in 0+1 dimensions - that describes uncompactified M-theory in 11 dimensions assuming that we study the large $N$ limit of the model with the $U(N)$ symmetry. As myself and later Susskind determined, one may also directly interpret the finite $N$ BFSS matrix model as describing ...

5

Matrix string theory http://arxiv.org/abs/hep-th/9701025 http://arxiv.org/abs/hep-th/9702187 http://arxiv.org/abs/hep-th/9703030 is indeed an exact description of fundamental type IIA strings (and similarly $E_8\times E_8$ heterotic strings) at any (e.g. weak) coupling where you can explicitly see the off-diagonal degrees of freedom. You could say ...

4

First of all, regardless of speculative comments in the original BFSS paper, different superselection sectors – different parts of the Hilbert space specified by the "background" (at least at infinity) – require different matrix model descriptions. The matrix model is known for $T^p$ with any radii; for $p\leq 3$, the theory is a super Yang-Mills theory on ...

2

The Christoffel–Darboux formula is not an asymptotic (in the sense of $N$ going to infinity) result, whereas the semicircle is valid for random matrices of infinite size. For finite matrices you obtain the oscillations you've got. To see this check out and plot formula (97) in http://arxiv.org/abs/math-ph/0412017 As to the traceless GUE, I'm no expert but ...

1

In the first action the $A_{\mu}$ are Hermitian. In the second action the $A_{\mu}$ are anti-Hermitian since we let $A_{\mu}\to\frac{i}{g}A_{\mu}$. The commutator of anti-Hermitian matrices are also anti-Hermitian. If we have $\text{Tr}(M^{2})$ , with $M$ being anti-Hermitian, then we can write it as $\text{Tr}(M^{2})=-\text{Tr}((iM)^{2})$ , with $iM$ ...

1

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 ...

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