General Relativity/Rigorous Definition of Tensors

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

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

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

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

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

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

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

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

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

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 ${\displaystyle (n,m)}$ and be able to insert either 1-forms or vectors into any slot of a given tensor.