General Relativity/Rigorous Definition of Tensors

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We have seen that a 1-form ("covariant vector") can be thought of an operator with one slot in which we insert a vector ("contravariant vector") and get the scalar $\mathit{\sigma }\left(𝐯\right)$. Similarly, a vector can be thought of as an operator with one slot in which we can insert a 1-form to obtain the scalar $𝐯\left(\mathit{\sigma }\right)$. As operators, they are linear, i.e., $\mathit{\sigma }(\alpha 𝐮+\beta 𝐯)=\alpha \mathit{\sigma }\left(𝐮\right)+\beta \mathit{\sigma }\left(𝐯\right)$.

A tensor of rank n is an operator with n slots for inserting vectors or 1-forms, which, when all n slots are filled, returns a scalar. In order for such an operator to be a tensor, it must be linear in each slot and obey certain transformation rules (more on this later). An example of a rank 2 tensor is $𝐓=T_{\nu }^{\mu }{𝐞}_{\mu }\otimes 𝐝x^{\nu }$. The symbol $\otimes$ (pronounced "tensor") tells you which slot each index acts on. This tensor $𝐓$ is said to be of type $(1,1)$ because it has one contravariant slot and one covariant slot. Since ${𝐞}_{\mu }$ acts on the first slot and $𝐝x^{\nu }$ acts on the second slot, we must insert a 1-form in the first slot and a vector in the second slot (remember, 1-forms act on vectors and vice-versa). Filling both of these slots, say with $\mathit{\sigma }$ and $𝐮$, will return the scalar $𝐓(\mathit{\sigma },𝐮)$. We can use linearity (remember, the tensor is linear in each slot) to evaluate this number:

$𝐓(\mathit{\sigma },𝐮)=T_{\nu }^{\mu }{𝐞}_{\mu }\otimes 𝐝x^{\nu }({\sigma }_{\alpha }𝐝x^{\alpha },u^{\beta }{𝐞}_{\beta })=T_{\nu }^{\mu }{𝐞}_{\mu }\left({\sigma }_{\alpha }𝐝x^{\alpha }\right)𝐝x^{\nu }\left(u^{\beta }{𝐞}_{\beta }\right)=T_{\nu }^{\mu }{\sigma }_{\alpha }{u}^{\beta }{\delta }_{\mu }^{\alpha }{\delta }_{\beta }^{\nu }=T_{\nu }^{\mu }{\sigma }_{\mu }{u}^{\nu }$

We don't have to fill all of the slots. This will of course not produce a scalar, but it will lower the rank of the tensor. For example, if we fill the second slot of $𝐓$, but not the first, we get a rank 1 tensor of type $(1,0)$ (which is a contravariant vector):

$𝐓(\cdot ,𝐮)=T_{\nu }^{\mu }{𝐞}_{\mu }\otimes 𝐝x^{\nu }(\cdot ,u^{\gamma }{𝐞}_{\gamma })=T_{\nu }^{\mu }{𝐞}_{\mu }(\cdot )𝐝x^{\nu }\left(u^{\gamma }{𝐞}_{\gamma }\right)=T_{\nu }^{\mu }{𝐞}_{\mu }{u}^{\gamma }{\delta }_{\gamma }^{\nu }=T_{\nu }^{\mu }{u}^{\nu }{𝐞}_{\mu }$

For another example, consider the rank 5 tensor $𝐒=S_{\beta \gamma }^{\alpha \mu \nu }{𝐞}_{\alpha }\otimes 𝐝x^{\beta }\otimes 𝐝x^{\gamma }\otimes {𝐞}_{\mu }\otimes {𝐞}_{\nu }$. This is a tensor of type $(3,2)$. We can fill all of its slots to get a scalar:

$𝐒(\mathit{\sigma },𝐮,𝐯,\mathit{\rho },\mathit{\xi })=S_{\beta \gamma }^{\alpha \mu \nu }{\sigma }_{\alpha }{u}^{\beta }{v}^{\gamma }{\rho }_{\mu }{\xi }_{\nu }$

Filling only the 3rd and 4th slots, we get a rank 3 tensor of type $(2,1)$:

$𝐒(\cdot ,\cdot ,𝐯,\mathit{\rho },\cdot )=S_{\beta \gamma }^{\alpha \mu \nu }{v}^{\gamma }{\rho }_{\mu }{𝐞}_{\alpha }\otimes 𝐝x^{\beta }\otimes {𝐞}_{\nu }$

As a final note, it should be mentioned that in General Relativity we will always have a special tensor called the "metric tensor" which will allow us to convert contravariant indices to covariant indices and vice-versa. This way, we can change the tensor type $(n,m)$ and be able to insert either 1-forms or vectors into any slot of a given tensor.