# Proof of result involving Pfaffian of a matrix

Show that $$\text{Pf} MAM^T = \text{det}M \cdot \text{Pf} A$$ for any matrix $M$ and antisymmetric $A$.

Attempt: $$\text{Pf} MAM^T = \frac{1}{2^N N!} \epsilon_{\alpha_1 \dots \alpha_{2N}} (MAM^T)_{\alpha_1 \alpha_2} \dots (MAM^T)_{\alpha_{2N-1} \alpha_{2N}} = \frac{1}{2^N N!}\epsilon_{\alpha_1 \dots \alpha_{2N}} M_{\alpha_1 \sigma_1} A_{\sigma_1 \delta_1} (M^T)_{\delta_1 \alpha_2} \dots M_{\alpha_{2N-1} \sigma_{2N-1}} A_{\sigma_{2N-1} \delta_{2N-1}} (M^T)_{\delta_{2N-1} \alpha_{2N}}$$ while $$\text{det} M = \epsilon_{\beta_1 \dots \beta_{2N}} M_{1, \beta_1} \dots M_{2N, \beta_{2N}}$$ and $$\text{Pf}A = \frac{1}{2^N N!} \epsilon_{\gamma_1 \dots \gamma_{2N}} (A)_{\gamma_1 \gamma_2} \dots (A)_{\gamma_{2N-1} \gamma_{2N}}$$

Working with the terms on the r.hs I see that $$\text{Pf}A \cdot \det M = \frac{1}{2^N N!} \epsilon_{\beta_1 \dots \beta_{2N}} \epsilon_{\gamma_1 \dots \gamma_{2N}} M_{1, \beta_1} \dots M_{2N, \beta_{2N}}(A)_{\gamma_1 \gamma_2} \dots (A)_{\gamma_{2N-1} \gamma_{2N}}$$ I don't see a way to proceed - is there perhaps another definition of $det$ I should use or can I argue based on these diagrammatic forms below?

• by definition $\text{pf}(B^T A B)^2=\det(B^T A B)=\det(B^T)\det(A)\det(B)=\det(B)^2 \det(A)$ using again the definition of the pfaffian u are done – tired Mar 29 '16 at 12:53
• Thanks for your response but I should have said I am not allowed to use the fact that $(\text{Pf}A)^2 = \det A$ – CAF Mar 29 '16 at 12:54
• I know there are a diagrammatic forms for $\text{Pf}A$ etc so maybe these would help? – CAF Mar 29 '16 at 12:57
• i guess ur problem boils down to "index confusion" so why not first proof the cases for N=1,2 and then try to generalize?\ – tired Mar 29 '16 at 13:00
• Yes I think it is because I have too many epsilons on each side. For $N=1$ I get $$\text{Pf}MAM^T = \frac{1}{2} \epsilon_{\alpha_1 \alpha_2} M_{\alpha_1 \beta_1} A_{\beta_1 \gamma_1} M_{\alpha_2 \gamma_1}$$ while for $$\text{det M} \text{Pf}A = \frac{1}{2} \epsilon_{\sigma_1 \sigma_2} A_{\sigma_1 \sigma_2} M_{1 a} M_{2b} \epsilon_{ab}$$ – CAF Mar 29 '16 at 13:10

Here is an approach using (possibly complex) Grassmann variables and Berezin integration$^1$.

1. Define the Pfaffian of a (possibly complex) antisymmetric matrix $A^{ij}=-A^{ji}$ (in $n$ dimensions$^2$) as $${\rm Pf}(A)~:=~\int \!d\theta_n \ldots d\theta_1~ e^{\frac{1}{2}\theta_i A^{ij}\theta_j} ~\stackrel{(5)}{=}~\frac{\partial}{\partial \theta_n} \ldots \frac{\partial}{\partial \theta_1} e^{\frac{1}{2}\theta_i A^{ij}\theta_j} ~=~\frac{1}{n!}\epsilon_{i_1\ldots i_n} \frac{\partial}{\partial \theta_{i_n}} \ldots \frac{\partial}{\partial \theta_{i_1}} e^{\frac{1}{2}\theta_i A^{ij}\theta_j}. \tag{1}$$

2. If we make a change of coordinates $$\theta^{\prime}_j~=~\theta_i M^i{}_j,\tag{2}$$ the chain rule becomes $$\frac{\partial}{\partial \theta_i}~=~M^i{}_j\frac{\partial}{\partial \theta^{\prime}_j} .\tag{3}$$

3. Therefore OP's first equation follows from \begin{align} {\rm Pf}(MAM^T) &~~\stackrel{(1)}{=}~\frac{1}{n!}\epsilon_{i_1\ldots i_n} \frac{\partial}{\partial \theta_{i_n}} \ldots \frac{\partial}{\partial \theta_{i_1}} e^{\frac{1}{2}\theta_i M^i{}_k A^{k\ell}M^j{}_{\ell}\theta_j}\cr &\stackrel{(2)+(3)}{=}~\frac{1}{n!}\epsilon_{i_1\ldots i_n}M^{i_1}{}_{j_1}\ldots M^{i_n}{}_{j_n} \frac{\partial}{\partial \theta^{\prime}_{j_n}} \ldots \frac{\partial}{\partial \theta^{\prime}_{j_1}} e^{\frac{1}{2}\theta^{\prime}_i A^{ij}\theta^{\prime}_j} \cr &~~=~\epsilon_{i_1\ldots i_n}M^{i_1}{}_{1}\ldots M^{i_n}{}_{n} ~\frac{\partial}{\partial \theta^{\prime}_{n}} \ldots \frac{\partial}{\partial \theta^{\prime}_{1}} e^{\frac{1}{2}\theta^{\prime}_i A^{ij}\theta^{\prime}_j} \cr &~~\stackrel{(1)}{=}~{\rm Det}(M)~{\rm Pf}(A).\end{align}\tag{4} $\Box$

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$^1$ We use the sign convention that Berezin integration $$\int d\theta_i~\equiv~\frac{\partial}{\partial \theta_i}\tag{5}$$ is the same as differentiation wrt. $\theta_i$ acting from left. See e.g. this Phys.SE post.

$^2$ One may show that the Pfaffian vanishes in odd dimensions.