# Why is lowering i in $A_{jk}^{i}=A(E_{i},\alpha,\beta)$ not $A_{ijk}=A(\sharp\omega^{i},\omega^{j},\omega^{k})$

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.

• I think you have it backwards. $A^i_{jk}$ are the components of a tensor that eats one covector and two vectors. Commented Feb 28 at 3:24
• 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 $A = A^i_{jk} E_i \otimes \omega^j \otimes \omega^k$. Commented Feb 28 at 3:30

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