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  The alpha-derivations

+ 1 like - 1 dislike
1167 views

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, 2021 in Mathematics by Antoine Balan (-80 points) [ revision history ]
edited Oct 17, 2021 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

+ 0 like - 0 dislike

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 Arnold Neumaier (15,787 points) [ revision history ]

That answer has no sense.

+ 0 like - 1 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, 2021 by Antoine Balan (-80 points) [ no revision ]

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