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From Paul Renteln, "Manifolds, Tensors and Forms" in a chapter on tensor fields:

Exercise 3.22 Not every object with indices is a tensor field. Let $X = X^i \partial / \partial x^i$ be a vector field. Show that the functions $\partial X^i/\partial x^j$ cannot be the components of any tensor field, by showing that they do not obey the transformation law for tensor field components.

I don't see why the given functions should not define a tensor field... the vector field $X$ and the coordinate functions $x^i$ are given, so $\partial X^i/\partial x^j$ are smooth functions for every $(i, j)$.

What wrong with letting $$T^{ij} := \frac{\partial X^i}{\partial x^j}$$ and $$T := T^{ij}\ e_i \otimes e_j = \frac{\partial X^i}{\partial x^j}\ e_i \otimes e_j$$ for some basis $\{e_i\}$? The components $T^{ij}$ are smoothly varying functions, so by definition $T$ should be a tensor field... ?

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    $\begingroup$ Where do you verify the definition? You seem to be using the definition involving a certain behaviour under coordinate transformations. You need to check this. $\endgroup$ – Thomas Apr 8 '14 at 19:30
  • $\begingroup$ @Thomas: by definition, "a tensor field ... is a map from $M$ to $T^r_s$ such that, on any coordinate patch, the components are smoothly varying functions". Since, in the exercise, the components come from a vector field and there is only a single coordinate patch, I think they are smooth? $\endgroup$ – googol analytics Apr 8 '14 at 19:44
  • $\begingroup$ No. You should review the definition. For a tensor, a coordinate change has to obey a certain transformation rule. $\endgroup$ – Thomas Apr 8 '14 at 19:46
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What you defined is a perfectly good tensor field. The exercise in the book is a little misleading. I think what the author really intended to say was something like this:

Let $X$ be a vector field. Show that there is no tensor field $T$ whose coefficients in every local coordinate chart $(x^1,\dots,x^n)$ are of the form $T^{ij} = \partial X^i/\partial x^j$.

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