Can anyone explain how to correctly carry out a derivative of a Grassmann exponential. Consider $e^{\theta}$ where $\theta$ is Grassmannian. Since, $\theta^2 = 0$ we have $$e^{\theta} = 1+\theta$$ And, hence, $$\frac{d}{d\theta}e^{\theta} = 1$$ which is obviously incorrect. Can someone please suggest the correct way to implement differentiation of a Grassmann exponential?
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Why would that be incorrect? Since $e^{\theta}=1$ at $\theta=0$, and the derivative of $e^{\theta}$ at $\theta=0$ (and everywhere else) is $1$, and all higher derivatives are $0$... you've shown that $e^{\theta}$ is linear (of course), and in fact is equal to $1+\theta$. Everything you wrote is consistent. – mjqxxxx Nov 09 '23 at 04:11
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@mjqxxxx Are you saying that Grassmann derivatives automatically implement derivation at $\theta = 0$? – Dr. user44690 Nov 09 '23 at 04:26
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The grassmannian tag refers to something entirely different. – Ted Shifrin Nov 09 '23 at 04:35
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Can you perhaps enlighten us as to what "grassmannian" mean in your context? All I know about grassmanian is the Grassmannian manifold. Perhaps $\theta$ refers to an element of the tangent space of a Grassmannian manifold? – Sangchul Lee Nov 09 '23 at 06:16
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@SangchulLee Sorry, I meant https://en.wikipedia.org/wiki/Grassmann_number. – Dr. user44690 Nov 09 '23 at 06:39
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I'm saying you're right that $(d/d\theta) e^{\theta}=1$. Indeed, the derivative of any function of a single Grassmann variable is a constant. And I'm asking why you think it's "obviously incorrect". – mjqxxxx Nov 09 '23 at 19:57