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  The alpha-derivations

+ 1 like - 0 dislike
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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?

asked Oct 17 in Mathematics by Antoine Balan (550 points) [ revision history ]
edited Oct 17 by Antoine Balan

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.

2 Answers

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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 by Arnold Neumaier (15,608 points) [ revision history ]

That answer has no sense.

+ 0 like - 0 dislike

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 by Antoine Balan (550 points) [ no revision ]

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