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Vít Tuček
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Consider the associativity for $fZ$ for any $f\in \mathcal{C}^\infty(M)$. Then on the right hand side of the associativty condition you are differentiating $f$ twice, while on the left hand side you are differentiating only once. Therefore such a connection cannot existsexist.

(This is pretty much what Sebastian is saying in his answer. I just thought it might be worthwhile to get rid of any notation and assumptions and explain the heart of the matter.)

Consider the associativity for $fZ$ for any $f\in \mathcal{C}^\infty(M)$. Then on the right hand side of the associativty condition you are differentiating $f$ twice, while on the left hand side you are differentiating only once. Therefore such a connection cannot exists.

(This is pretty much what Sebastian is saying in his answer. I just thought it might be worthwhile to get rid of any notation and assumptions and explain the heart of the matter.)

Consider the associativity for $fZ$ for any $f\in \mathcal{C}^\infty(M)$. Then on the right hand side of the associativty condition you are differentiating $f$ twice, while on the left hand side you are differentiating only once. Therefore such a connection cannot exist.

(This is pretty much what Sebastian is saying in his answer. I just thought it might be worthwhile to get rid of any notation and assumptions and explain the heart of the matter.)

Source Link
Vít Tuček
  • 8.6k
  • 2
  • 30
  • 61

Consider the associativity for $fZ$ for any $f\in \mathcal{C}^\infty(M)$. Then on the right hand side of the associativty condition you are differentiating $f$ twice, while on the left hand side you are differentiating only once. Therefore such a connection cannot exists.

(This is pretty much what Sebastian is saying in his answer. I just thought it might be worthwhile to get rid of any notation and assumptions and explain the heart of the matter.)