< General Relativity

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 $\mathbf {\sigma } \left(\mathbf {v} \right)$. 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 $\mathbf {v} \left(\mathbf {\sigma } \right)$. As operators, they are linear, i.e., $\mathbf {\sigma } \left(\alpha \mathbf {u} +\beta \mathbf {v} \right)=\alpha \mathbf {\sigma } \left(\mathbf {u} \right)+\beta \mathbf {\sigma } \left(\mathbf {v} \right)$.

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 $\mathbf {T} =T_{\ \nu }^{\mu }\mathbf {e} _{\mu }\otimes \mathbf {d} x^{\nu }$. The symbol $\otimes$ (pronounced "tensor") tells you which slot each index acts on. This tensor $\mathbf {T}$ is said to be of type $(1,1)$ because it has one contravariant slot and one covariant slot. Since $\mathbf {e} _{\mu }$ acts on the first slot and $\mathbf {d} x^{\nu }$ 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 $\mathbf {\sigma }$ and $\mathbf {u}$, will return the scalar $\mathbf {T} \left(\mathbf {\sigma } ,\mathbf {u} \right)$. We can use linearity (remember, the tensor is linear in each slot) to evaluate this number:

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

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 $\mathbf {T}$, but not the first, we get a rank 1 tensor of type $(1,0)$ (which is a contravariant vector):

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

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

$\mathbf {S} \left(\mathbf {\sigma } ,\mathbf {u} ,\mathbf {v} ,\mathbf {\rho } ,\mathbf {\xi } \right)=S_{\ \beta \gamma }^{\alpha \ \ \mu \nu }\sigma _{\alpha }u^{\beta }v^{\gamma }\rho _{\mu }\xi _{\nu }$
Filling only the 3rd and 4th slots, we get a rank 3 tensor of type $(2,1)$:

$\mathbf {S} \left(\cdot \ ,\cdot \ ,\mathbf {v} ,\mathbf {\rho } ,\cdot \right)=S_{\ \beta \gamma }^{\alpha \ \ \mu \nu }v^{\gamma }\rho _{\mu }\mathbf {e} _{\alpha }\otimes \mathbf {d} x^{\beta }\otimes \mathbf {e} _{\nu }$

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.