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I'm learning/playing around with tensors and somehow got this contradiction, suppose $\{v_i\}$ and $\{w_i\}$ are basis for a vector space $V$ and $\{ v^i \}$ and $\{w^i\}$ are basis for the dual space $V^\ast $ with $v^i v_j = \delta ^i _j $, then
$$ \begin{aligned} \delta^i_j &= v^i v_j \\ &= \left(\sum_{l=1}^n v^i (w_l) w^l \right) \left(\sum_{k=1}^n w^k (v_j) w_k \right) &&\text{by change of basis} \\ &= \sum_{l=1}^n \sum_{k=1}^n v^i(w_l) \ w^k(v_j) \ w^l w_k &&\text{since $v^i(w_l)$ and $w^k(v_j)$ are scalars} \\ &= \sum_{l=1}^n \sum_{k=1}^nv^i(w_l) \ w^k(v_j) \ \delta^l_k &&\text{by $w^l w_k = \delta^l_k $ being a dual basis} \\ &= \sum_{l=1}^n v^i(w_l) \ w^l(v_j) &&\text{applied $\delta^l_k$} \\ &= v^i \left( \sum_{l=1}^nw_l w^l \right) v_j &&\text{by associativity } \\ &= v^i n v_j &&\text{since $\textstyle w_l w^l = \delta^l_l = \sum_{l=1}^n 1=n$} \\ &= n \delta ^i_j \end{aligned} $$ so it implies $\delta^i_j = 0 $ which is definitely not correct, so my question is, where is the mistake?

Edit: fixed dummy variable and used explicit sums.

Note, change of basis: suppose $v^i = d_l w^l$ for some $d_j \in \mathbb R$, then
$$ \begin{aligned} v^i(w_j) &= (d_l w^l ) (w_j) \\ &= d_l \delta^l_j \\ &= d_j \\ \Rightarrow \ v^i &= (v^i(w_l) ) w^l. \end{aligned} $$

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    $\begingroup$ hint: you have too many j's in the 2nd row. $\endgroup$
    – hyportnex
    Aug 8, 2022 at 14:16
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    $\begingroup$ @hyportnex That's not the problem. I submitted an edit to make the equations clear. $\endgroup$
    – Luessiaw
    Aug 8, 2022 at 14:19
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    $\begingroup$ In OP's notation $w_j w^j$ is not a scalar. $w^j(w_j)$ is and equals to $\delta^j_j$. But in general habits there is no difference. $\endgroup$
    – Luessiaw
    Aug 8, 2022 at 14:22

3 Answers 3

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Choose another set of bases $e_\alpha$. $\boldsymbol v_i,\boldsymbol w_j$ are all vectors and have components along them. That is, $$ \boldsymbol v^{i}=v^{i}_\alpha\boldsymbol e^\alpha,\quad \boldsymbol w^{i}=w^{i}_\alpha\boldsymbol e^\alpha,\\ \boldsymbol v_{i}=v_{i}^\alpha\boldsymbol e_\alpha,\quad \boldsymbol w_{i}=w_{i}^\alpha\boldsymbol e_\alpha, $$ where vectors are in bold, distinguished by the superscribpt $i$, and their components are in normal italic, marked by Greek characters $\alpha$.

The normalization relationship is $$ \delta^\alpha_\beta=\boldsymbol e^\alpha(\boldsymbol e_\beta). $$

$\boldsymbol v$'s form a bases iff $$ \delta^i_j= \boldsymbol v^i(\boldsymbol v_j) =v^i_\alpha\boldsymbol e^\alpha (v^\beta_j\boldsymbol e_\beta) =v^i_\alpha v^\beta_j\delta^\alpha_\beta=v^i_\alpha v^\alpha_j, $$ and similar to $\boldsymbol w$'s.

Suppose basis change is in the form $$ \boldsymbol v^i=C^i_j\boldsymbol w^j \Leftrightarrow v^i_\alpha\boldsymbol e^\alpha =C^i_jw^j_\alpha\boldsymbol e^\alpha\Leftrightarrow v^i_\alpha = C^i_jw^j_\alpha. $$ Multyiply both sides of the above equation by $w^\alpha_k$ and apply the normalization of $\boldsymbol w$'s, we get $$ C^i_k = v^i_\alpha w^\alpha_k. $$ Therefore $$ \boldsymbol v^i=v^i_\alpha w^\alpha_j \boldsymbol w^j=\boldsymbol v^i(\boldsymbol w_j)\boldsymbol w^j. $$ There's no much difference from OP's formula, but we should write the components, $$ v^i _\beta = v^i_\alpha w^\alpha_j w^j_\beta. $$ They are all numbers and obey associative and commutative laws. Therefore in OP's fifth line, we're faced with $$ \boldsymbol v^i(\boldsymbol w_l)\boldsymbol w^l(\boldsymbol v_j) =v^i_\alpha w_l^\alpha w^l_\beta v_j^\beta =v^i_\alpha v_j^\beta w^l_\beta w_l^\alpha. $$ Pay attention to the super- and subscripts, $$ w^l_\alpha w_l^\beta\not=w^l_\alpha w^\alpha_l=\delta^l_l=n. $$ So what is $w^l_\alpha w_l^\beta$? As you may guess, it is $\delta^\alpha_\beta$. Let's check.

Since $\boldsymbol w$'s and $\boldsymbol e$'s are two sets of bases, any vector $\boldsymbol A$ can be represented by $$ \boldsymbol A=A^\alpha \boldsymbol e_\alpha=A^i\boldsymbol w_i, $$ where $A^\alpha$ and $A^i$ are different numbers. Apply $\boldsymbol w^j$ to both sides, we get $$ A^j=\boldsymbol w^j(\boldsymbol e_\alpha) A^\alpha=w^j_\alpha A^\alpha. $$ The above equation implies, $w^j_\alpha$ is the transfromation matrix for this basis change. Meanwhile basis change conserves inner product of vectors, so $$ \boldsymbol A\cdot\boldsymbol B= A^\alpha B_\alpha =A^l B_l=w^l_\alpha w_l^\beta A^\alpha B_\beta. $$ Because $\boldsymbol A,\boldsymbol B$ are arbirary, we get $$ w^l_\alpha w^\beta_l=\delta^\alpha_\beta. $$ Therefore $$ \boldsymbol v^i(\boldsymbol w_l)\boldsymbol w^l(\boldsymbol v_j) =v^i_\alpha v_j^\beta w^l_\beta w_l^\alpha =v^i_\alpha v_j^\beta \delta^\alpha_\beta=v^i_\alpha v_j^\alpha = \delta ^i_j. $$ Everything goes fine.

The point is, I think, a vector $\boldsymbol v$ is a one-order tensor, $v_\alpha$. But when we talk about a set of vectors, $\{\boldsymbol v_i\}$, they form a two-order tensor, $v^j_\alpha$. The product of two vectors is in general a four-order tensor, $v^i_\alpha v_j^\beta=\delta^i_j\delta_\alpha^\beta$ in this case. Contract $i$ with $j$ or $\alpha$ with $\beta$ we get a two-order tensor, and contract all variables we get a scalar.

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First, you can shorten things considerably by not changing coordinates and then changing back on the second term.

$\delta^i_j=v^iv_j=(v^i(w_l)w^l)v_j=v^i(w_lw^l)v_j=v^inv_j=n\delta^i_j$

We can also drop the now-fixed $v_j$ on the terms in the middle.

$v^i=(v^i(w_l)w^l)=v^i(w_lw^l)=v^in$

And now we can see better what is going on.

Let's put it in vector language, as that might be a bit more intuitively familiar to a learner.

$\vec{v}=(\vec{v}\cdot\vec{x})\vec{x}+(\vec{v}\cdot\vec{y})\vec{y}=\vec{v}(\vec{x}\cdot\vec{x})+\vec{v}(\vec{y}\cdot\vec{y})=\vec{v}(\vec{x}\cdot\vec{x}+\vec{y}\cdot\vec{y})=2\vec{v}$

The dot product is not associative with the scalar product.

And the change of basis doesn't look quite right to me. If we're trying to do $\vec{v}=(\vec{v}\cdot\vec{x})\vec{x}+(\vec{v}\cdot\vec{y})\vec{y}$ then the $v$ has to be contracted with the $w$, not the $w$ with itself. It should look something like $v^i=v^j(w_j)w^i$ ($=v^j\delta^i_j$).

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  • $\begingroup$ I thought associativity works because they are all tensor compositions, hence I could re-bracket them. I've added the derivation for my change of basis in the question edit. I do agree with you that the notation is really bad and $w_l w^l$ should really equal one and not summed over (?). $\endgroup$
    – Tony.Y
    Aug 9, 2022 at 1:45
  • $\begingroup$ In your example, $w_1=\vec x$ and $w_2=\vec y$, and the contraction happens to $w$'s, not $w$ with $v$. $\endgroup$
    – Luessiaw
    Aug 9, 2022 at 2:59
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In the sixth line $w_lw^l=\delta_l^l$ seems not to be ture, because it is not a scalar but a covector left multiplied by a vector. The scalar is $w^l(w_l)$ and it equals to $\delta^l_l$.

Noting that $v^i(w_l)w^l=v^i$, by change of basis. This implies $w_lw^l=1$. Then the derivation returns to the origin, $$ v^i(w_l)w^l(v_j)=v^i(v_j)=\delta^i_j. $$

But this conflicts with what I'm used to. Things need to be checked. To my knowledeg, $w_iw^j$ and $w^jw_i$ are often regarded as the same quantity.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Aug 8, 2022 at 15:01
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    $\begingroup$ $w_lw^l$ is not $1$. Sure, by appropriate usage of double duality and abuse of notation, this can be seen as the identity mapping on the vector space $V$, but I don't think such overload of notation is helpful here. Anyway, you're right that if done correctly, we end up where we started. $\endgroup$
    – peek-a-boo
    Aug 8, 2022 at 15:32
  • $\begingroup$ Is there a way we could know when $w_l w^l = 1$ and when $w_l w^l = \delta^l_l = n$ ?because the notations are the same (?) and by tensor associativity, I should be able to re-bracket them, (but in this case not?). $\endgroup$
    – Tony.Y
    Aug 9, 2022 at 1:51

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