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In the solution to one exercise, we have the identification $$R_{\partial_i,\partial_j}\partial_k\Big|_p = R^m_{ijk}(p) \partial_{m}\Big|_p$$ for the riemann-curvature tensor with respect to a point $p \in M$, M SR manifold, and coordinate-vector-fields $\partial_i,\partial_j,\partial_k \in \mathfrak{X}(M)$. Now, if I try to show this explicitly, we get $$R^m_{ijk} = dx^m(R_{\partial_i,\partial_j}\partial_k) = \ldots = \partial_{i}\Gamma^m_{jk}+\Gamma^p_{jk}\Gamma^{m}_{ip}-\partial_{j}\Gamma^m_{ik}-\Gamma^{p}_{ik}\Gamma^{m}_{jp}$$

Multiplying by $\partial_{m}$ we get $$R^{m}_{ijk} \partial_{m} = \partial_{i}\Gamma^m_{jk}\partial_{m}+\Gamma^p_{jk}\Gamma^{m}_{ip} \partial_{m}-\partial_{j}\Gamma^m_{ik} \partial_{m}-\Gamma^{p}_{ik}\Gamma^{m}_{jp} \partial_{m} $$

Okay, so far so good. Now, if we compute $R_{\partial_i,\partial_j}\partial_{k}$ we get $$R_{\partial_i,\partial_j}\partial_{k} = \nabla_{\partial_i}\nabla_{\partial_j}\partial_k-\nabla_{\partial_j}\nabla_{\partial_i}\partial_k \quad (\text{since} \quad [\partial_i,\partial_j] = 0)$$ $$=\nabla_{\partial_i}(\Gamma^p_{jk} \partial_p)-\nabla_{\partial_j}(\Gamma^p_{ik} \partial_p)$$ $$=\partial_i(\Gamma^p_{jk})\partial_p+\Gamma^{p}_{jk}\nabla_{\partial_i}\partial_p-\partial_{j}(\Gamma^{p}_{ik})\partial_p-\Gamma^{p}_{ik}\nabla_{\partial_j}\partial_{p}$$$$= \partial_i(\Gamma^{p}_{jk})\partial_p+\Gamma^{p}_{jk}\Gamma^{c}_{ip}\partial_c-\partial_j(\Gamma^{p}_{ik})\partial_p-\Gamma^{p}_{ik}\Gamma^{d}_{jp} \partial_d$$

Now, in the last step, I presume I am allowed to change the first term $$\partial_i(\Gamma^{p}_{jk})\partial_p$$ to $$\partial_i(\Gamma^{m}_{jk})\partial_{m}$$ without changing the indices from $p$ to $m$ in the second term $$\Gamma^{p}_{jk}\Gamma^{c}_{ip}\partial_c?$$ Because then, I can proceed, setting $c = m$ in the second term, $p=m$ in the third term (without changing $p$ for the other terms) and finally $d=m$ to find the identity I want to show (se prelude). So my main question is, am I allowed to change $p$ individually? And is there a simpler way to show this identity?

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    $\begingroup$ Yes you are allowed. Each sum occurs individualy and as such the different terms don't care about what you name the summation indices. The only instance in which you may pay attention is if you replace the name of an index by another index that's involved in the multiplication. For example, $p$ to $k$ in the first term would be wrong because that would imply a second summation that wasn't there originally $\endgroup$ Aug 19, 2023 at 19:54

2 Answers 2

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Is there a simpler way to show this identity?

Yes: In your proof you used the fact that \begin{equation}\tag{1} R{}^m_{ijk} = dx{}^m(R_{\partial_i,\partial_j}\partial_k). \end{equation} Let $U$ be the domain of $x$, then \begin{equation}\tag{2} \forall X\in\mathfrak{X}(U)=dx{}^mX\partial_m. \end{equation} You are simply considering the case $X=R_{\partial_i,\partial_j}\partial_k$: \begin{equation} R_{\partial_i,\partial_j}\partial_k\overset{(2)}{=}(dx{}^mR_{\partial_i,\partial_j}\partial_k)\partial_m\overset{(1)}{=}R^m_{ijk}\partial _m \end{equation}

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The short answer to your second question is that there is nothing to show. The definition of the components of a tensor is $$T_{ijk...}^{lmn...}:= T(e_i,e_j,e_k,...,\epsilon^l,\epsilon^m,\epsilon^n,...)$$ where the $e$'s are the basis vectors and the $\epsilon$'s are the basis covectors. So when you write

$$R_{\partial_i,\partial_j}\partial_k\Big|_p = R^m_{ijk}(p) \partial_{m}\Big|_p$$

This is already just a definition.
Now, presumably you first defined the Riemann tensor by its components, so this ends up being an "identity", but, really, it is backwards.

As for the actual calculation of the components, yours is as simple as it gets

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  • $\begingroup$ There might be different conventions, but in my course the covectors would come first in the expression you have. I don´t really know what you mean. Since an $(r,s)$-tensor takes r arguments from the dual-space $V^*$ of a vector space $V$ first and then $s$ arguments from the vector space itself. $\endgroup$
    – Ben123
    Aug 19, 2023 at 20:13
  • $\begingroup$ The order of covectors of course doesn't matter. And would you clarify what you don't understand in my answer? Yes, a tensor does all those things, but I don't get what that has to do with what I wrote $\endgroup$ Aug 19, 2023 at 20:34
  • $\begingroup$ I would disagree, it does matter, then I would specify the domain and codomain of the tensor. Otherwise it is unclear that a student understands how a tensor functions. Clearly there is something to show, since in the solution our TA writes that we need a specific lemma to show this identity. If there was nothing to show, then why would one use the lemma? The lemma in question is precisely the lemma of which I gave a proof of above. $\endgroup$
    – Ben123
    Aug 19, 2023 at 20:38
  • $\begingroup$ But thanks for your input, I modtly eanted to make sure that this was fine, it is possible I am missing something obvious. $\endgroup$
    – Ben123
    Aug 19, 2023 at 21:11

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