# Describe $f$ in terms of the tensor products of $\alpha^i$ and $\alpha^j$. Is this inner product? What are its coefficients?

My book is An Introduction to Manifolds by Loring W. Tu. In an exercise, we are asked to describe a bilinear function in terms of tensor products.

The answer is given at the back is $$f= \sum g_{ij} \alpha^i \otimes \alpha^j$$.

What is the connection between the following example in the text and the above exercise?

Does this just follow from an earlier example on inner product? Or do they just both follow from an even earlier example that tells us $$v^i = \alpha^i(v)$$?

I think the example was meant to be moved to an exercise or vice-versa.

Also, I have seen four solutions to this exercise, and not one makes a reference to the example.

To clarify,

1. Is $$f$$ in the exercise is the same as the inner product as in the example?

2. In the exercise, is $$g_{ij} = f(e_i, e_j)$$ ?

• I think so because $$f(e_l, e_m) = \sum_{i,j} g_{ij} (e_l)^i (e_m)^j = \sum_{i,j} g_{ij} (e_l)^i (e_m)^j = \sum_{i,j} g_{ij} \alpha^i(e_l) \alpha^j(e_m) = \sum_{i,j} g_{ij} \delta^i_l \delta^j_m = g_{lm}$$
• You're right that the example solves the exercise. Worse oversights occur in textbooks :) – Ted Shifrin Feb 28 at 1:41
• @TedShifrin I edited my question. To clarify, 1. Is $f$ in the exercise is the same as the inner product as in the example? 2. In the exercise, is $g_{ij} = f(e_i, e_j)$ ? Thanks! – Selene Auckland Feb 28 at 10:36
• $f$ is more general than inner product, of course. Note it needn't even be a symmetric bilinear form. But yes, $f(e_i,e_j)=g_{ij}$. – Ted Shifrin Feb 28 at 16:01
• @TedShifrin Thanks! – Selene Auckland Mar 2 at 7:51