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Diffstat (limited to 'topology.tex')
| -rw-r--r-- | topology.tex | 101 |
1 files changed, 63 insertions, 38 deletions
diff --git a/topology.tex b/topology.tex index daacab2..6570ca4 100644 --- a/topology.tex +++ b/topology.tex @@ -6,6 +6,7 @@ \usepackage[bitstream-charter]{mathdesign} \usepackage[dvipsnames]{xcolor} \usepackage{anyfontsize,authblk} +\usepackage{xfrac} % sfrac \urlstyle{sf} @@ -448,10 +449,10 @@ make use of the identity |\det A| &=\lim_{\epsilon\to0}\frac{(\det A)^2}{\sqrt{\det(A+i\epsilon I)}\sqrt{\det(A-i\epsilon I)}} \\ - &=\frac1{(2\pi)^N}\int d\bar{\pmb\eta}_1\,d\pmb\eta_1\,d\bar{\pmb\eta}_2\,d\pmb\eta_2\,d\mathbf a_+\,d\mathbf a_-\, + &=\frac1{(2\pi)^N}\int d\bar{\pmb\eta}_1\,d\pmb\eta_1\,d\bar{\pmb\eta}_2\,d\pmb\eta_2\,d\mathbf a_+\,d\mathbf a_2\, e^{ -\bar{\pmb\eta}_1^TA\pmb\eta_1-\bar{\pmb\eta}_2^TA\pmb\eta_2 - -\frac12\mathbf a_+^T(A+i\epsilon I)\mathbf a_+-\frac12\mathbf a_-^T(A-i\epsilon I)\mathbf a_- + -\frac12\mathbf a_+^T(A+i\epsilon I)\mathbf a_+-\frac12\mathbf a_2^T(A-i\epsilon I)\mathbf a_2 } \end{aligned} \end{equation} @@ -1082,18 +1083,18 @@ Starting from \eqref{eq:abs.kac-rice}, we make the substitution &\big|\det\partial\partial L(\mathbf x,\pmb\omega)\big| =\lim_{\epsilon\to0}\frac1{(2\pi)^{N+M+1}}\int d\bar{\pmb\eta}_1\,d\pmb\eta_1\,d\bar{\pmb\eta}_2\,d\pmb\eta_2\, d\bar{\pmb\gamma}_1\,d\pmb\gamma_1\,d\bar{\pmb\gamma}_2\,d\pmb\gamma_2\, - d\mathbf a_+\,d\mathbf a_-\,d\mathbf b_+\,d\mathbf b_-\, \notag \\ + d\mathbf a_1\,d\mathbf a_2\,d\mathbf b_1\,d\mathbf b_2\, \notag \\ &\qquad\times\exp\Bigg\{ - -\frac12\begin{bmatrix}\mathbf a_+^T&\mathbf b_+^T\end{bmatrix}\big(\partial\partial L(\mathbf x,\pmb\omega)+i\epsilon I\big)\begin{bmatrix}\mathbf a_+\\\mathbf b_+\end{bmatrix} - -\frac12\begin{bmatrix}\mathbf a_-^T&\mathbf b_-^T\end{bmatrix}\big(\partial\partial L(\mathbf x,\pmb\omega)-i\epsilon I\big)\begin{bmatrix}\mathbf a_-\\\mathbf b_-\end{bmatrix} + -\frac12\begin{bmatrix}\mathbf a_1^T&\mathbf b_1^T\end{bmatrix}\big(\partial\partial L(\mathbf x,\pmb\omega)+i\epsilon I\big)\begin{bmatrix}\mathbf a_1\\\mathbf b_1\end{bmatrix} + -\frac12\begin{bmatrix}\mathbf a_2^T&\mathbf b_2^T\end{bmatrix}\big(\partial\partial L(\mathbf x,\pmb\omega)-i\epsilon I\big)\begin{bmatrix}\mathbf a_2\\\mathbf b_2\end{bmatrix} \notag \\ &\hspace{6em}-\begin{bmatrix}\bar{\pmb\eta}_1^T&\bar{\pmb\gamma}_1^T\end{bmatrix}\partial\partial L(\mathbf x,\pmb\omega)\begin{bmatrix}\pmb\eta_1\\\pmb\gamma_1\end{bmatrix} -\begin{bmatrix}\bar{\pmb\eta}_2^T&\bar{\pmb\gamma}_2^T\end{bmatrix}\partial\partial L(\mathbf x,\pmb\omega)\begin{bmatrix}\pmb\eta_2\\\pmb\gamma_2\end{bmatrix} \Bigg\} \label{eq:abs.det.exp} \end{align} -where the $\mathbf a_\pm$ are in $\mathbb R^N$, the $\mathbf b_\pm$ are in -$\mathbb R^M$, and the $\pmb\eta$ and $\pmb\gamma$ are Grassmann vectors just as in +where the $\mathbf a_{\sfrac12}$ are in $\mathbb R^N$, the $\mathbf b_{\sfrac12}$ are in +$\mathbb R^M$, and the $\pmb\eta_{\sfrac12}$ and $\pmb\gamma_{\sfrac12}$ are Grassmann vectors just as in \eqref{eq:det.exp}, except with an extra copy of each. This zoo of vectors quickly becomes tiring. Thankfully, there is a way to compactly represent this calculation again using superspace vectors. @@ -1104,16 +1105,16 @@ Consider the vectors &=\mathbf x +\bar\theta_1\pmb\eta_1+\bar{\pmb\eta}_1\theta_1\bar\theta_2\theta_2 +\bar\theta_2\pmb\eta_2+\bar{\pmb\eta}_2\theta_2\bar\theta_1\theta_1 \\ - &\qquad+\frac1{\sqrt2}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)\mathbf a_+ - +\frac i{\sqrt2}(\bar\theta_1\theta_1+\bar\theta_2\theta_2)\mathbf a_- + &\qquad+\frac1{\sqrt2}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)\mathbf a_1 + +\frac i{\sqrt2}(\bar\theta_1\theta_1+\bar\theta_2\theta_2)\mathbf a_2 +\bar\theta_1\theta_1\bar\theta_2\theta_2i\hat{\mathbf x} \notag \\ \sigma_k(1,2) &=\omega_k +\bar\theta_1\gamma_{1k}+\bar{\gamma}_{1k}\theta_1\bar\theta_2\theta_2 +\bar\theta_2\gamma_{2k}+\bar{\gamma}_{2k}\theta_2\bar\theta_1\theta_1 \\ - &\qquad+\frac1{\sqrt2}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)b_{-k} - +\frac i{\sqrt2}(\bar\theta_1\theta_1+\bar\theta_2\theta_2)b_{+k} + &\qquad+\frac1{\sqrt2}(\bar\theta_1\theta_2+\bar\theta_2\theta_1)b_{1k} + +\frac i{\sqrt2}(\bar\theta_1\theta_1+\bar\theta_2\theta_2)b_{2k} +\bar\theta_1\theta_1\bar\theta_2\theta_2\hat\omega_k \notag \end{align} @@ -1125,7 +1126,7 @@ form =\lim_{\epsilon\to0}\int d\pmb\phi\,d\pmb\sigma\,e^{ \int d1\,d2\,L(\pmb\phi(1,2),\pmb\sigma(1,2)) -\frac{i\epsilon}2 - (\|\mathbf a_+\|^2-\|\mathbf a_-\|^2+\|\mathbf b_+\|^2-\|\mathbf b_-\|^2) + (\|\mathbf a_1\|^2-\|\mathbf a_2\|^2+\|\mathbf b_1\|^2-\|\mathbf b_2\|^2) } \end{equation} Note that unlike in the derivation of the average Euler characteristic, @@ -1151,49 +1152,42 @@ Following similar steps as for the Euler characteristic, we first take the avera We once again have a Gaussian integral in the parameters inside the Lagrange multipliers $\sigma_k$ for $k=1,\ldots,M$. However, here we must expand the superfields at this stage. Before that, we introduce order parameters analogous -to before. Alongside those in \eqref{eq:ops.bos} and \eqref{eq:ops.ferm}, we also define +to before. Alongside those in \eqref{eq:ops.bos}, we have \begin{align} - A_{++}=\frac{\mathbf a_+\cdot\mathbf a_+}N + A_{ij}=\frac{\mathbf a_i\cdot\mathbf a_j}N && - A_{--}=\frac{\mathbf a_-\cdot\mathbf a_-}N + X_i=\frac{\mathbf x\cdot\mathbf a_i}N && - A_{+-}=\frac{\mathbf a_+\cdot\mathbf a_-}N - && - X_+=\frac{\mathbf x\cdot\mathbf a_+}N - && - X_-=\frac{\mathbf x\cdot\mathbf a_-}N - \\ - \hat X_+=-i\frac{\hat{\mathbf x}\cdot\mathbf a_+}N - && - \hat X_-=-i\frac{\hat{\mathbf x}\cdot\mathbf a_-}N + \hat X_i=-i\frac{\hat{\mathbf x}\cdot\mathbf a_i}N && H_{ij}=\frac{\bar{\pmb\eta}_i\cdot\pmb\eta_j}N && J=\frac{\pmb\eta_1\cdot\pmb\eta_2}N && - \bar J=\frac{\bar{\pmb\eta}_1\cdot\bar{\pmb\eta}_2}N + m_i = \frac{\mathbf x_0\cdot\mathbf a_i}N \end{align} +where the $i$ and $j$ run over $1$ and $2$. These come with a volume term in the action \begin{equation} \begin{aligned} \mathcal V=\frac12\log\det\left( \begin{bmatrix} - C & iR & X_+ & X_- \\ - iR & D & i\hat X_+ & i\hat X_- \\ - X_+ & i\hat X_+ & A_{++} & A_{+-} \\ - X_- & i\hat X_- & A_{+-} & A_{--} + C & iR & X_1 & X_2 \\ + iR & D & i\hat X_1 & i\hat X_2 \\ + X_1 & i\hat X_1 & A_{11} & A_{12} \\ + X_2 & i\hat X_2 & A_{12} & A_{22} \end{bmatrix} -\begin{bmatrix} m \\ i\hat m \\ - m_+ \\ - m_- + m_1 \\ + m_2 \end{bmatrix} \begin{bmatrix} m & i\hat m & - m_+ & - m_- + m_1 & + m_2 \end{bmatrix} \right) \qquad \\ -\frac12\log\det\begin{bmatrix} @@ -1207,9 +1201,9 @@ These come with a volume term in the action As before, we can immediately integrate out $\sigma_0$, which fixes certain of the order parameters. In particular, we find \begin{align} C=1 && - X_+=0 && - X_-=0 && - G_+=-R-\frac12(A_{++}+A_{--}) + X_1=0 && + X_2=0 && + G_+=-R-\frac12A_+ \end{align} where we have defined the symmetric and antisymmetric combinations of $G_{11}$ and $G_{22}$ \begin{align} @@ -1249,11 +1243,42 @@ and likewise a Gaussian integral in the Grassmann variables with kernel The effective action then has the form \begin{equation} \mathcal S - =-\frac\alpha2\left( + =\hat m+\frac12i\epsilon A_--\frac\alpha2\left( \log\det K-\log\det K'+V_0^2(K^{-1})_{22} \right)+\mathcal V \end{equation} - +It is not helpful to write out this entire expression, which is quite large. +However, we only find saddle points of this action with $\bar +J=J=G_{12}=G_{21}=G_-=\hat X_1=\hat X_2=m_1=m_2=A_{12}=0$. Setting these variables to zero, we find +\begin{align} + \notag + \mathcal S + =\hat m + +\frac12i\epsilon A_- + -\alpha\frac12\log\frac{ + f'(1)(Df(1)+R^2f'(1))-\frac12((R-\frac12A_+)^2+\frac12 A_-^2)f(1)f''(1) + }{ + \frac14(R+\frac12A_+)^2f'(1)^2 + } + \\ + \notag + -\alpha\frac12\log\frac{ + \epsilon^2+i\epsilon f'(1)A_-+\frac14(A_+^2-A_-^2)f'(1)^2 + }{ + \frac14(R+\frac12A_+)^2f'(1)^2 + } + +\frac12\log\frac{D(1-m^2)+R^2-2Rm\hat m+\hat m^2}{\frac14(R+\frac12A_+)^2} + \\ + +\frac12\log\frac{A_+^2-A_-^2}{(R+\frac12A_+)^2} + -\frac12\alpha V_0^2\left( + f(1)+\frac{R^2f'(1)^2}{Df'(1)-\frac12((R-\frac12A_+)^2+\frac12A_-^2)f''(1)} + \right)^{-1} +\end{align} +One solution to these equations at $\epsilon=0$ is $A_-=0$ and $A_+=2R^*$ with +$D=\hat m=0$ and $R=R^*$, exactly as for the Euler characteristic. The resulting effective +action as a function of $m$ is also exactly the same. We find two other +solutions that differ from this one, but with formulae too complex to share in +the text. One alternate solution has $A_-=0$ and one has $A_-\neq0$. \section{The quenched shattering energy in {\oldstylenums 1}\textsc{frsb} models} |