# The alpha-derivations

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Let $M$ be a differential manifold and $\alpha \in {\bf R}$. I define an $\alpha$-derivation $X_{\alpha}$ as an application of the smooth functions such that:

$$X_{\alpha} (fg)=X_{\alpha}(f)g^{\alpha}+f^{\alpha}X_{\alpha}(g)$$

$$X_{\alpha}(f+g)^{1/\alpha}=X_{\alpha}(f)^{1/\alpha}+X_{\alpha}(g)^{1/\alpha}$$

The definition makes sense because:

$$X_{\alpha}((fg)h)=X_{\alpha}(f(gh))=X_{\alpha}(fgh)=$$

$$=X_{\alpha}(f)(gh)^{\alpha}+f^{\alpha}X_{\alpha}(g)h^{\alpha}+(fg)^{\alpha}X_{\alpha}(h)$$

What are the $\alpha$-derivations of the real smooth functions?

edited Oct 17, 2021

Is X(f) supposed to be positive? Otherwise one has problems interpreting the power.

We may suppose that f and X(f) are positiv in the formulas. We could also take alpha to be an odd integer, or even integer and X(f) positiv, or rational with odd primes, in these cases the powers are defined I think.

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If $x^\alpha:=\alpha(X)$ is a smooth bijection then the general solution is $$X_\alpha(f):=\delta(f)d\alpha(f)/df$$ for some derivation $\delta$. Proof by composition with $\alpha$ (or its inverse).

answered Oct 18, 2021 by (15,767 points)

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We can calculate $X_{\alpha}(f(f+x))$ in two ways as $X_{\alpha}(f(f+x))=X_{\alpha}(f^2+xf)$

$$X_{\alpha}(f(f+x))=(f^{\alpha}+(f+x)^{\alpha})X_{\alpha}(f)$$

$$X_{\alpha}(f^2 + xf)^{1/\alpha}= (2^{1/\alpha}f + x)X_{\alpha}(f)^{1/\alpha}$$

We obtain $\alpha=1$ or $X_{\alpha}(f)=0$.

answered Oct 19, 2021 by (-80 points)

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