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If $g_t$ is a time dependent metric on $M$ and $\nabla$ is connection, if $Y$ is a smooth vector field on $M$, then why does $\nabla_{Y} \partial_t g_t$ makes sense?

I don't understand why $\partial_t g_t$ is a tensor.

Namely, I know that if $T$ is a $(a,b)$ Tensor defined on $M$, and $Y$ is a smooth vector field on $M$, then $\nabla_{Y}T$ makes sense and has an explicit formula.

FD_bfa
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    Because it satisfies the definition of a tensor. Could you be more specific about which part of the definition you’re not sure about? – Deane May 12 '23 at 23:34
  • What type of tensor is it @Deane ? –  May 12 '23 at 23:36
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    $\partial_tg_t$ is a twice-covariant symmetric tensor field. Symmetry and multilinearity of $g$ are properties which are linear in the variable $g$, and therefore are preserved under taking derivatives. But in general, $\partial_tg_t$ is no longer positive-definite (it may even be degenerate). – Ivo Terek May 12 '23 at 23:40
  • @IvoTerek Sorry, could you write a proof as to why it is a twice-covariant symmetric tensor field? I am confused about the definitions. Is the time partial derivative of a time dependent tensor always a tensor? If you write that as an answer ill accept. Thnx –  May 12 '23 at 23:43
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    This seems like a worthwhile exercise. – Deane May 12 '23 at 23:55
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    At least pointwisely, for each $x\in M$, $g_t(x)$ is an element in a finite dimensional vector space. So $t\mapsto g_t(x)$ is a curve in that vector space. So...... $\partial_t g_t (x)$ is also an element in that vector space. – Arctic Char May 13 '23 at 11:58
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    See my answer on PhySE Is a perturbation of a tensor field a tensor field? – peek-a-boo Jun 04 '23 at 20:08
  • @peek-a-boo exactly what I needed. Upvoted. –  Jun 04 '23 at 21:32

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