# What is not a tensor? [duplicate]

I've been taking GR, and all of a sudden I am not sure that I know the necessary and sufficient requirements of the tensor coefficients. This is because the lecturer asked me to prove that that "there exist no tensor constructed from the metric and its derivatives, other than the metric". This leads me to have exactly two questions:

(1) Suppose $g_{\mu \nu}$ is the metric tensor, $c$ is a real number and $\nabla_\rho$ the covariant derivative. Why are the following not tensors?

• $c g_{\mu \nu}$
• $\nabla_\rho g_{\mu \nu}$

(2) Is it wrong that the sufficient and necessary condition for the tensor coefficient is that they have well defined transformation rules, such that they takes the same form in all coordinates? Is this not the same as saying that the tensor transforms as a tensor?

• $\cos(g_{\mu\nu})$ obiously does not transform in the right way, and the covariant derivative of the metric vanishes. Add "non-trivial" to the original statement if you want. – fqq Mar 9 '17 at 12:44
• The statement "there exist no tensor constructed from the metric and its derivatives, other than the metric" is a strange and wrong statement, the Riemann tensor of the Levi-Civita connection can be expressed in terms of the second derivatives of the metric. Also, why do you suspect your second statement is wrong - what is your definition of a tensor, after all? – ACuriousMind Mar 9 '17 at 12:57
• $cg_{\mu\nu}$ and $\nabla_\rho g_{\mu\nu}$ certainly are tensors. – Ryan Unger Mar 9 '17 at 19:40