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Dimensional reduction of Yang-Mills to m(atrix) theory

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The Yang-Mills action are usually given by $$S= \int\text{d}^{10}\sigma\,\text{Tr}\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\theta^{T}\gamma^{\mu}D_{\mu}\theta\right)$$

with the field strength defined as $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}-ig\left[A_{\mu},A_{\nu}\right]$ , $A_{\mu}$ being a U(N) Hermitian gauge field in the adjoint representation, $\theta$ being a $16\times1$ Majorana-Weyl spinor of $SO(9)$ in the adjoint representation and $\mu=0,\dots,9$ . The covariant derivative is given by $D_{\mu}\theta=\partial_{t}\theta-ig\left[A_{\mu},\theta\right]$. We are using a metric with mostly positive signs.
 

We re-scale the fields by $A_{\mu}\to\frac{i}{g}A_{\mu}$ and let $g^{2}\to\lambda$ which gives us $$S=\int\text{d}^{10}\sigma\,\text{Tr}\left(\frac{1}{4\lambda}F_{\mu\nu}F^{\mu\nu}-\theta^{T}\gamma^{\mu}D_{\mu}\theta\right)$$ with the field strength defined as $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+\left[A_{\mu},A_{\nu}\right]$ and the covariant derivative $D_{\mu}\theta=\partial_{t}\theta+\left[A_{\mu},\theta\right]$.

Now we perform a dimensional reduction from $9+1$ to $0+1$ , so that all the fields only depend on time, thous all spatial derivatives vanish i.e. $\partial_{a}(\text{Anything})=0$ . The $10$ -dimensional vector field decomposes into $9$ scalar fields $A_{a}$ which we rename $X^{a}$ and one gauge field $A_{0}$ which we rename $A$ . This gives (note that $\gamma^{t}=\mathbb{I}$ and that $\gamma^{a}=\gamma_{a}$. $$F_{0a}= \partial_{t}X^{a}+\left[A,X^{a}\right],\quad F_{ab}=+\left[X^{a},X^{b}\right] \gamma^{t}D_{t}\theta= \partial_{t}\theta+\left[A,\theta\right],\quad\gamma^{a}D_{a}\theta=\gamma_{a}\left[X^{a},\theta\right]$$

The action for this theory is then $$S=\int\text{d}t\,\text{Tr}\left(\frac{1}{2\lambda}\bigg\{-\left(D_{t}X^{a}\right)^{2}+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}\bigg\}-\theta^{T}D_{t}\theta-\theta^{T}\gamma_{a}\left[X^{a},\theta\right]\right)$$ with the covariant derivative defined as $D_{t}X^{a}=\partial_{t}X^{a}+\left[A,X^{a}\right]$ and $D_{t}\theta=\partial_{t}\theta+\left[A,\theta\right]$

Now to the question. I need the potential energy $V=+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}$ to be negative, not positive.

Taylor has a discussion on this in his paper "Lectures on D-branes, Gauge Theory and M(atrices)" (http://arxiv.org/abs/hep-th/9801182) on page 10, where he writes:
"Because the metric we are using has a mostly positive signature, the kinetic terms have a single raised 0 index corresponding to a change of sign, so the kinetic terms indeed have the correct sign. The commutator term $\left[X^{a},X^{b}\right]^{2}$ which acts as a potential term is actually negative definite. This follows from the fact that $\left[X^{a},X^{b}\right]^{\dagger}=\left[X^{b},X^{a}\right]=-\left[X^{a},X^{b}\right]$. Thus, as expected, kinetic terms in the action are positive while potential terms are negative."

But I don't understand where the Hermitian conjugate $^\dagger$ comes from, to me this term is just: $$\left[X^{a},X^{b}\right]^{2}=\left[X^{a},X^{b}\right]\left[X_{a},X_{b}\right]$$

Note that Taylor uses a little different conventions when he re-scales, instead of $A_{\mu}\to\frac{i}{g}A_{\mu}$ he uses $A_{\mu}\to\frac{1}{g}A_{\mu}$ and $\theta \to\frac{1}{g}\theta$. But this should not cause any troubles I think.


This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Natanael

asked Nov 10, 2013 in Theoretical Physics by Natanael (75 points) [ revision history ]
retagged May 21, 2014 by dimension10

1 Answer

+ 2 like - 0 dislike

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 $iM$ being Hermitian. Since the eigenvalues of an Hermitian matrix is real and we take the trace of the square it, it follows that $\text{Tr}(M^{2})\leq0$ . Changing the anti-Hermitian matrices to Hermitian matrices changes the sign.

This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Natanael
answered Nov 22, 2013 by Natanael (75 points) [ no revision ]

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