Proving an object is a tensor by contraction

Question

I have seen the claim that if $V_{\mu}$ is a covector, and $T^{\mu\nu}V_{\nu}$ is a vector, then $T^{\mu\nu}$ is a tensor. I am trying to prove this, and I have two questions.

It seems to rest upon the statement that $$(T^{\mu\nu}V_{\nu})'=T'^{\mu\nu}V'_{\nu}.$$ If $T$ is a tensor, this is clearly true. Is such a 'linearity' also true for a generic $T$? If not, is there an example where this is not true that does not involve derivatives?

Also, why does $\partial^{\mu}\nabla^{\nu}$ not falsify this statement? It is not a tensor, yet its contraction with $V_{\nu}$ is.

EDIT: An example with less ambiguity is $\partial^{\mu}W^{\nu}V_{\nu}$ for some vector $W^{\mu}$. This is a vector, but we would then draw the false conclusion that $\partial^{\mu}W^{\nu}$ is a tensor.

Also to clarify, my question is not about how to prove this theorem. My question is about i) why the the equation I wrote out for co-ordinate transformation holds in general, given that we have no idea what kind of object $T$ is, and ii) why my example does not disprove the theorem.

EDIT2: Here is a proof that $\nabla_{\mu}V^{\mu}$ is a scalar, so that we can move past this and address my question. Either you use that we are contracting two vectors, and so it is clearly a scalar since the equation must have homogeneous weight. Or we can do it 'properly': Let $$J^{\mu}_{\nu} = \frac{\partial x^{\mu}}{\partial x'^{\nu}}.$$ Then $$\partial_{\sigma} \text{log}\det J = \partial_{\sigma}\text{Tr}\text{log}J = \text{Tr}\partial_{\sigma}\text{log}J =\text{Tr} J^{-1}\partial_{\sigma}J = 0$$ since $$\partial_{\sigma}J^{\mu}_{\nu} = \frac{\partial}{\partial x'^{\nu}}\frac{\partial x^{\mu}}{\partial x^{\sigma}} = \frac{\partial}{\partial x'^{\nu}} \delta^{\mu}_{\sigma}= 0.$$ Similarly $$ \frac{\partial x^{\sigma}}{\partial x'^{\mu}}\partial_{\sigma}\frac{\partial x'^{\mu}}{\partial x^{\nu}}=0,$$ again just by the chain rule, commutativity of derivatives, and the fact that the Kronecker delta is constant. It is now trivial to see that \begin{align*} \frac{1}{\sqrt{g}}\partial_{\mu}(\sqrt{g}V^{\mu}) &\rightarrow\frac{\partial x^{\sigma}}{\partial x'^{\mu}}\frac{1}{\text{det}J\sqrt{g}}\partial_{\sigma}(\text{det}J \sqrt{g}\frac{\partial x'^{\mu}}{\partial x^{\nu}}V^{\nu})\\ &=\frac{1}{\sqrt{g}}\partial_{\mu}(\sqrt{g}V^{\mu}). \end{align*}

Answer 1

Why not simply write out how a vector transforms, how tensor transforms and do it by linear algebra.

Let $S:\{x^\alpha\}^{\alpha=1\dots N}$ be one set of coordinates and $\tilde{S}:\{x^\alpha\}^{\alpha=1\dots N}$ be another set.

If $V_\alpha$ is a co-vector in $S$ whilst $\tilde{V}_\alpha$ is a co-vector in $\tilde{S}$, then, by definition:

$V_\alpha = \frac{\partial \tilde{x}^\beta}{\partial x^\alpha}\tilde{V}_\beta$, and, inverting this $\tilde{V}_\mu = \frac{\partial x^\beta}{\partial \tilde{x}^\mu}V_\beta$

Similarly if $T^{\alpha\beta}V_\beta$ is a vector then:

$\left(T^{\alpha\beta}V_\beta\right)=\frac{\partial x^\alpha}{\partial \tilde{x}^\gamma}\tilde{\left(T^{\gamma\mu}V_\mu\right)}=\frac{\partial x^\alpha}{\partial \tilde{x}^\gamma}\left(\tilde{T}^{\gamma\mu}\tilde{V}_\mu\right)=\frac{\partial x^\alpha}{\partial \tilde{x}^\gamma}\tilde{T}^{\gamma\mu}\frac{\partial x^\beta}{\partial \tilde{x}^\mu}V_\beta$

If this is true for all $V_\alpha$ in co-vector space, your result should be easy to prove.

Regarding the $\nabla^\nu V_\nu$. Do it properly, assuming Levi-Civita connection:

$\nabla^\nu V_\nu = g^{\nu\mu}\nabla_\mu V_\nu=\nabla_\mu g^{\nu\mu} V_\nu=\nabla_\mu V^\mu = \partial_\mu V^\mu + \Gamma^\eta_{\mu\eta}V^\mu=\partial_\mu V^\mu + \frac{\partial_\mu g}{2g} V^\mu=\frac{1}{\sqrt{g}}\partial_\mu\left(\sqrt{g} V^\mu\right)$

This is not necessarily a true scalar, because determinant of the metric is not a scalar (it is more like a density that picks up jacobian determinant when one changes coordinates). Hence partial derivative of this is not necessarily a vector/co-vector. See Rund & Lovelock.

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Following comments. I repeat. Do it properly. Let's look at the weight.

$\phi=\frac{1}{\sqrt{g}}\partial_\nu\left(\sqrt{g}V^\nu\right)$

in coordinates $S$. What would it be in $\tilde{S}$? Let's do it bit by bit. Firstly $g=J^2\cdot\tilde{g}$, where $J=\left|\frac{\partial\left(\tilde{x}\right)}{\partial\left(x\right)}\right|$

Then:

$\frac{1}{\sqrt{g}}\partial_\nu\left(\sqrt{g}V^\nu\right)=\frac{1}{J\cdot\sqrt{\tilde{g}}}\cdot\frac{\partial \tilde{x}^\kappa}{\partial x^\nu}\tilde{\partial}_\kappa \left(J\frac{\partial x^\nu}{\partial \tilde{x}^\alpha}\sqrt{\tilde{g}} \tilde{V}^\alpha\right) $

So:

$\phi=\frac{1}{\sqrt{g}}\partial_\nu\left(\sqrt{g}V^\nu\right) = \frac{\tilde{\partial}_\alpha J}{J}\tilde{V}^\alpha + \left(\partial_\alpha V^\alpha -\tilde{\partial}_\alpha \tilde{V}^\alpha\right) + \frac{1}{\sqrt{\tilde{g}}}\cdot\tilde{\partial}_\mu \left(\sqrt{\tilde{g}} \tilde{V}^\mu\right)\neq\tilde{\phi}$

Where one would expect:

$\tilde{\phi}=\frac{1}{\sqrt{\tilde{g}}}\cdot\tilde{\partial}_\mu \left(\sqrt{\tilde{g}} \tilde{V}^\mu\right)$

Unless you can make the first two terms disappear, I maintain that $\phi$ is not a true scalar.

Similar arguments go for the other bit. Show your work, then we could try to find the problem.

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Yet another addition

Ok, lets back up. You have a co-vector $V_\mu$ and a vector $U^\nu=T^{\nu\mu}V_\mu$, where $T^{\mu\nu}$ is simply a linear transformation that works in coordinates $S$.

Question 1. Does this hold for any co-vector $V_\mu$, i.e. is $U^\nu=T^{\nu\mu}V_\mu$ a vector for all co-vectors?

Question 2. You then proceed to talk about $\tilde{T}^{\nu\mu}$ which is $T'^{\nu\mu}$ in your notation. What is that object?

You said '''the definition of $\tilde{T}$ is whatever $T$ looks like in the new co-ordinate system'''. This is too vague. There are clear rules for transforming tensors between different coordinate frames, but everything else needs additional work. Currently your main question is like this ''Lets say I have A and B. How do I proove that A=B, but without defining what is A?'' I got around this by incorporating definition into the derivation. You did not like this. So what is your definition of $\tilde{T}$

Answer 2

Let

$$V=V_{\mu} dx^{\mu}$$ be the co-vector, and let

$$T=T^{\mu\nu} \frac{\partial}{\partial x^{\mu}} \otimes\frac{\partial}{\partial x^{\nu}}$$

be the tensor.

And let the tensor product of $T$ and $V$ be

$$T^{'}=T^{\mu\nu}V_{\nu} \frac{\partial}{\partial x^{\mu}} \otimes\frac{\partial}{\partial x^{\nu}}\otimes dx^{\nu}$$.

Contraction (see "https://en.wikipedia.org/wiki/Tensor_contraction") is summing over $2$ like indices, which in your case, you would leave you with a vector

$$T^{'}=T^{\mu}\frac{\partial}{\partial x^{\mu}}$$

after contraction - which may appear to work at a singe point on the manifold.

But you still haven't proved $T=T^{\mu\nu} \frac{\partial}{\partial x^{\mu}} \otimes\frac{\partial}{\partial x^{\nu}}$ is a tensor.

There's only $1$ way to show that $T$ is tensor, and that's to show it transforms like a tensor under a change of coordinates.

Answer 3

A tensor is just a multi linear function. So you just check for linearity that's it.

Edit: this is quite trivial. If after contraction it is linear in all arguments and contraction is linear its a tensor.