I am trying to understand how to lower an index of a (1,2)-tensor $A(v,\alpha,\beta)$. As far as I understand A eats a vector and two covectors. The coefficients are given by $A_{jk}^{i}$.

I want to lower $i$,i.e. the "vector slot". To my understanding by lowering $i$ i get a tensor that eats three covectors, say $A'(\gamma,\alpha,\beta)$.

Why is $A'(\omega^{i},\omega^{j},\omega^{k})=A(\sharp\omega^{i},\omega^{j},\omega^{k})=g^{pq}A(E_{q},\omega^{j},\omega^{k})$ wrong?

$\{\omega^{k}\}$ is the dual basis.

No matter what I do I don't get $A_{ijk}=g_{pq}A_{ij}^{q}$, i.e. the factor $g_{pq}$

Doesn't A need a vector and two covectors as input? In particular A needs a vector in the first slot.Therefore I thought I need to convert $\omega^{i}$ via the musical isomorphism $\sharp$ into a vector.

I had a look at Jeffrey Lee's book. There it seems to me my $A_{jk}^{i}$ is somehow $A_{i}^{jk}$.

Does it make sense to say: Lowering the number of contravariant slots is done by using $\sharp$ and lowering the number of covariant slots is done by using $\flat$?

I am very confused and would really apreciate any help. Many thanks in advance.

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    $\begingroup$ I think you have it backwards. $A^i_{jk}$ are the components of a tensor that eats one covector and two vectors. $\endgroup$ Commented Feb 28 at 3:24
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    $\begingroup$ If your $A$ really does eat $1$ vector and $2$ covectors, then $A = A^{jk}_i \omega^i \otimes E_j \otimes E_k$, so that $A(v,\alpha,\beta) = A^{jk}_i \omega^i(v) \alpha(E_j) \beta(E_k)$. In particular, it doesn't make sense to "lower the $i$ index" in this context. On the other hand, if you insist that $A$ has components $A^i_{jk}$, then (as Ted Shifrin said) $A$ has to eat $2$ vectors and $1$ covector, so that $A = A^i_{jk} E_i \otimes \omega^j \otimes \omega^k$. $\endgroup$ Commented Feb 28 at 3:30

1 Answer 1


It seems like the comments have this essentially wrapped up, but for the sake of clarity... When getting lost in index juggling I find that explicitly writing the basis elements can be very helpful. So for your tensor we have...

$$A_{jk}^{i}(e_{i}\otimes e^{j} \otimes e^k)$$

Where $\{e_i\}$ is our vector basis and $\{e^i\}$ is our dual basis.

So, as was mentioned in the comments, your tensor must "eat" 1 covector and 2 vectors in order to produce an element of the underlying field. If I wish, for whatever reason, to lower the "$i$" index then I can, indeed, apply the musical isomorphism so long as my vector space is finite dimensional and endowed with a proper non-degenerate bilinear form.

$$\therefore (A_{jk}^{i})^\flat=A_{jk}^{i}(g_{im}e^{m}\otimes e^{j} \otimes e^k)=g_{im}A_{jk}^{i}(e^{m}\otimes e^{j} \otimes e^k)=A_{mjk}$$


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