General Relativity/Rigorous Definition of Tensors

< 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 $$\mathbf{T} = T^\mu_{\ \nu} \mathbf{e}_\mu \otimes \mathbf{d}x^\nu$$. 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:

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

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):

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

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

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

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

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

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.