2
$\begingroup$

So I am trying to calculate the components $T^i\,_{j k}$ relative to a coordinate basis, of the torsion tensor defined as: $$ \bf{T(u,v)= \nabla_{u}v-\nabla_v u-[u,v]} $$

Or: $$ T= T^i\,_{j k} e_i\otimes\omega^j \otimes \omega^k $$ Now, I have read through a couple resources and found that I can find the components by evaluating my first expression on my basis vectors. I am curious why this is so and if every tensor is like this.

I should add that the $e_\alpha$ and $\omega^\beta$, are my coordinate vectors and the dual vectors (respectively).

$\endgroup$
2
$\begingroup$

If $\{e_i\}$ is a given frame then there are functions such that

$$\nabla_{e_i}e_j=\alpha_{ij}^ke_k,\quad [e_i,e_j]=c_{ij}^ke_k.$$

Then,

$$T(e_i,e_j)=\nabla_{e_i}e_j-\nabla_{e_j}e_i-[e_i,e_j]=\alpha_{ij}^ke_k-\alpha_{ji}^ke_k-c_{ij}^ke_k.$$ That is,

$$T=(\alpha_{ij}^k-\alpha_{ji}^k-c_{ij}^k)e^i\otimes e^j \otimes e_k.$$

Note that since $T$ is a tensor it is enough to compute it on a given basis (of course, this works for any tensor). If you want to evaluate it at $u=u^ie_i,v=v^je_j$ you have

$$T(u,v)=u^iv^jT(e_i,e_j),$$ since, as I have said, it is a tensor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.