# Confused about tensor notations of how vector and covectors act on each other

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}

• hint: you have too many j's in the 2nd row. Aug 8, 2022 at 14:16
• @hyportnex That's not the problem. I submitted an edit to make the equations clear. Aug 8, 2022 at 14:19
• 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. Aug 8, 2022 at 14:22

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.

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 $$v^i=v^j(w_j)w^i$$ ($$=v^j\delta^i_j$$).

• 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 (?). Aug 9, 2022 at 1:45
• In your example, $w_1=\vec x$ and $w_2=\vec y$, and the contraction happens to $w$'s, not $w$ with $v$. Aug 9, 2022 at 2:59

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.

• $w_lw^l$ 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. Aug 8, 2022 at 15:32
• 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?). Aug 9, 2022 at 1:51