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Hi,

I have a superspace spanned by 4 commuting coordinates + 2 anti-commuting ones $\{x^\mu,\theta^\alpha\}$, I have to do the change of coordinates $dx^\mu\to dy^\mu= dx^\mu+d\theta^\alpha \eta_\alpha^{\;\:\mu}$ where $\eta$ have to be a local function i.e: $\eta\equiv\eta(x)$, and leave the $d\theta$'s unchanged, so how can I translate this change on coordinates themselves? In particular I need to express $\partial_\mu$ in the new coordinate system.

If the $\eta$ fcts where global it would be simply $x^\mu\to y^\mu= x^\mu+\theta^\alpha \eta_\alpha^{\;\:\mu}$, so what if $\eta$'s where not global?

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Please do not use words like "local" and "global" in this context. By a "global function" you mean a constant and by a "local function" you simply mean a function. This sort of nonstandard language might be confusing to some people. –  José Figueroa-O'Farrill Feb 20 '10 at 14:46
    
Sorry, bad physicist language I know ;-) –  Pedro Feb 20 '10 at 16:17
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2 Answers 2

up vote 5 down vote accepted

First of all, I would not call $dx^\mu \mapsto dy^\mu$ a change of coordinates. Not every transformation on 1-forms is going to be the pullback by a smooth map. In fact, your question is whether the transformation $(dx^\mu,d\theta^\alpha) \mapsto (dy^\mu,d\theta^\alpha)$ comes from a change of coordinates.

Pullbacks commute with exterior derivative, so that if $dy^\mu$ is a pullback then applying $d$ to your $dy^\mu$ should give zero. If you do that you find that it is not unless $\eta$ is constant.

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Thank you for this very clear answer... nevertheless I wonder how I can now express my partial derivatives! –  Pedro Feb 20 '10 at 16:20
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The partial derivatives are canonically dual to the differentials, so they transform as the "inverse transpose". Be careful that just because you have declared that $d\theta^\alpha$ does not transform, this does not mean that the partial derivative $\partial/\partial\theta^\alpha$ does not either! –  José Figueroa-O'Farrill Feb 20 '10 at 17:43
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Ok here are the partial derivatives $k_M$:

they are dual to $\{b^M\} = \{dy^\mu,d\theta^\alpha\}$ ie:

$b^M(k_N)=\delta_N^M$

Writing $k_N=(k_N)^{\mu}\frac{\partial}{\partial x^\mu}+(k_N)^{\alpha}\frac{\partial}{\partial \theta^\alpha}$ and solving $b^\nu(k_\mu)=\delta^\nu_\mu,\;b^\mu(k_\alpha)=0; \; b^\alpha(k_\nu)=0;\;b^\alpha(k_\beta)=\delta^\alpha_\beta$ one finds:

$(k_\mu)^\nu=\delta^\nu_\mu,\;(k_\mu)^\alpha=0;\;(k_\alpha)^\beta=\delta^\beta_\alpha;\;(k_\alpha)^\nu=-\eta_\alpha^{\;\;\mu}$

So:

$\frac{\partial}{\partial y^\mu} = \frac{\partial}{\partial x^\mu}$

$\frac{\partial}{\partial y^\alpha} = -\eta^{\;\;\mu}_\alpha \frac{\partial}{\partial x^\mu} + \frac{\partial}{\partial \theta^\alpha}$

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There is some problem with the notation in the left-hand side of the last equation. You'd like to say that this is $\partial/\partial\theta^\alpha$, but then then equation would be inconsistent. Anyway, it's clear what you mean. –  José Figueroa-O'Farrill Feb 22 '10 at 17:52
    
Thank you ..... –  Pedro Mar 12 '10 at 16:32
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