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I'm studying about tensors and have already understood the following theorem:

$C^1_1$ is the unique linear function such that $C_1^1(v\otimes\eta)=\eta(v)=v(\eta)$ for all $v\in V$ and $\eta\in V^*$.

Now I'm reading about this theorem:

Let $1\leq k\leq r, 1\leq l\leq s$. Then there exists a unique linear map $C^k_l:\mathcal{T}^r_s(V)\to\mathcal{T}^{r-1}_{s-1}(V)$ such that $$C^k_l(v_1\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\eta^s)=v_k(\eta^l)v_1\otimes\ldots\otimes \hat v_k\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\hat\eta^l\otimes\ldots\otimes\eta^s$$ where the hat indicates that that term is omitted.

The proof given in the book is kind of unintuitive to me, so I tried an alternative method of proving it. Here's my attempt:

Let's define $$C^k_l(A)\equiv A^{i_1\ldots i_r}_{j_1\ldots j_s}\ C_1^1(h_{i_k}\otimes\theta^{j_l})\ h_{i_1}\otimes\ldots\otimes \hat h_{i_k}\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes\hat\theta^{j_l}\otimes\ldots\otimes\theta^{j_s}$$ where $h_i$ and $\theta^j$ are basis vectors and covectors respectively. With the above definition, $$C^k_l(v_1\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\eta^s)$$ $$=(v_1)^{i_1}\ldots (v_r)^{i_r}(\eta^1)_{j_1}\ldots (\eta^s)_{j_s}C_1^1(h_{i_k}\otimes\theta^{j_l})\ h_{i_1}\otimes\ldots\otimes \hat h_{i_k}\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes\hat\theta^{j_l}\otimes\ldots\otimes\theta^{j_s}$$ Pulling $(v_k)^{i_k}$ and $(\eta^l)_{j_l}$ into the $C_1^1$ argument and the rest of the components into the big tensor product on the right, we get the RHS of the desired result.

Uniqueness follows from the uniqueness of $C_1^1$.

Are there any issues with the above proof? Would appreciate any help!

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  • $\begingroup$ Anyone with some idea? $\endgroup$
    – Shirish
    Jan 14, 2022 at 12:20
  • $\begingroup$ Why is the proof given in the book not unintuitive? What book are you reading? $\endgroup$
    – user1010411
    Jan 17, 2022 at 15:04
  • $\begingroup$ When you say "let's define", you need to justify why it can be represented in that way and what the coefficients $A_J^I$ are. $\endgroup$
    – user1010411
    Jan 17, 2022 at 15:06
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    $\begingroup$ @ripples: Hi! The book is Semi-Riemannian Geometry by Stephen Newman. It was unintuitive at first, but it's fine now. However I still want to follow through with my proof on whether it's fine / how I can complete it. As for your questions: I know the way I defined it seems ad hoc, but I don't think it's wrong to specify any definition for $C^k_l$ as long as it's well-defined. And I think it's well-defined since $C^1_1$ is well-defined (due to the previous theorem I wrote at the question start). Also $A^I_J$ represent the components of $A$ in the (arbitrary) basis that we chose. $\endgroup$
    – Shirish
    Jan 17, 2022 at 15:26
  • $\begingroup$ I think a potential gap could be that the way I defined $C^k_l$ is basis dependent. So maybe I need to rephrase the definition to say that instead of an arbitrary basis, I define $C^k_l(A)$ to be of that form w.r.t. a particular basis. And then prove that $C^k_l(A)$ takes the same form w.r.t. any other basis. Though I'm not sure if that is essential or not. $\endgroup$
    – Shirish
    Jan 18, 2022 at 8:58

2 Answers 2

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Any linear map $C$ between two vector spaces ${\cal U}$ and ${\cal W}$ is uniquely defined as long as it is defined on a basis of ${\cal U}\,.$ A basis of ${\cal T}^r_s(V)$ is $$\tag{1} h_{i_1}\otimes\,...\,\otimes\,h_{i_r}\otimes \theta^{j_1}\,\otimes\,...\otimes\,\theta^{j_s} $$ where all $i_k$ and all $j_l$ run through $\{1,...,n\}$ and $\{h_1,...,h_n\}$ is a basis of $V$ and $\{\theta_1,...\theta_n\}$ is a basis of $V^*\,.$ Unfortunately I do not have access to Newman's book but I strongly believe that he defines the map $C^k_l$ on a larger set than a basis of ${\cal T}^r_s(V)$ by allowing each of the factors $v_i$ and $\eta^j$ to be a linear combination of the basis vectors $h_\mu\,,$ resp. $\theta_\nu\,.$ (If his $v_i$ were all different, or his $\eta^j\,,$ this would not define $C^k_l$ on a basis of ${\cal T}^r_s(V)\,.$ Clearly, in (1) the factors are not all different.)

The only thing (if anything) that needs to be checked is if the contraction form of the right hand side of \begin{align} &C^k_l(v_1\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\eta^s)\\&=v_k(\eta^l)\,v_1\otimes\ldots\otimes \hat v_k\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\hat\eta^l\otimes\ldots\otimes\eta^s\tag{2} \end{align} is compatible with the same form for basis vectors: \begin{align} &C^k_l(h_{i_1}\otimes\,...\,\otimes\,h_{i_r}\otimes \theta^{j_1}\,\otimes\,...\otimes\,\theta^{j_s})\\&=h_{i_k}(\theta^{j_l})\,h_{i_1}\otimes\ldots\otimes \hat h_{i_k}\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes\hat\theta^{j_l}\otimes\ldots\otimes\theta^{j_s}\,.\tag{3} \end{align} By linearity it is enough to expand only the hatted factors $$ v_k=\alpha{^\mu}_k\,h_\mu\,,\quad\eta^l={\beta^l}_\nu\,\theta^\nu\quad\text{ (using summation convention) } $$ and assume that in (2) the un-hatted factors are basis vectors. In other words we need to check that (3) implies \begin{align} &C^k_l(h_{i_1}\otimes\ldots\otimes v_k\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes \eta^l\otimes\ldots\otimes\theta^{j_s})\\&=v_k(\eta^l)\,h_{i_1}\otimes\ldots\otimes \hat{v}_k\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes \hat{\eta}^l\otimes\ldots\otimes\theta^{j_s}\,.\tag{2'} \end{align} From $$ v_k(\eta^l)=\alpha{^\mu}_k{\beta^l}_\nu\,h_\mu(\theta^\nu) $$ this should however by obvious. If not please note that $\mu,\nu$ are just dummy indices which we can replace: $$ v_k(\eta^l)=\alpha{^{i_k}}_k{\beta^l}_{j_l}\,h_{i_k}(\theta^{j_l})\,. $$

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    $\begingroup$ Thanks! So then my attempt at the proof that I showed in the question seems correct I think $\endgroup$
    – Shirish
    Jan 18, 2022 at 15:17
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    $\begingroup$ Looks so to me. $\endgroup$
    – Kurt G.
    Jan 18, 2022 at 17:35
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The existence part is fine: you basically define $C_l^k$ on a basis of $\mathcal{T}_s^r(V)$ in the way it is supposed to be and then extend it linearly.

The problem is in the "uniqueness" part. Unlike the book's proof, it does not directly follow from the uniqueness of $C_1^1$ in your proof.

What you need to show is that if $L$ is another linear map that satisfies the same condition for simple tensors, then $L$ is equal to the $C_l^k$ you define. (Note that $L$ is not necessarily defined via a basis.) But this is also trivial since $L$ and $C_l^k$ have the same value on the simple tensors, particularly the basis tensors you choose for $\mathcal{T}_s^r(V)$; so they are equal by linearity.


Your book (Semi-Riemannian geometry by Newman) defines $C_l^k$ rather differently. They do it in a "coordinate-free" way.

Given a tensor $A\in\mathcal{T}_s^r(V)$, they define explicitly how $C_l^k(A)$ acts on vectors and covectors. More specifically, given the following element in $V^{*r}\times V^s$: $$ (\eta^1,\cdots, \eta^{k-1},\eta^{k+1},\cdots,\eta^r,w_1,\cdots,w_{l-1},w_{l+1},\cdots,w_s)\;, $$ it is mapped by $C_l^k(A)$ to the (unique) $(1,1)$-contraction of the following tensor in $\mathcal{T}_1^1(V)$: $$ (\eta,w)\mapsto A(\eta^1,\cdots, \eta^{k-1},\eta,\eta^{k+1},\cdots,\eta^r,w_1,\cdots,w_{l-1},w,w_{l+1},\cdots,w_s)\;. $$

They then prove that the defined map $C_l^k$ has the desired property on the simple tensors.

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