What is the scalar product of tensors?

Question

Given there a vector space $V$ with a scalar product $g(v_1,v_2)$ on it, what is the scalar product on, say, $V \otimes V^*$ ?

According to Jeffrey Lee's "Manifolds and Differential Geometry" (see 7.6 Metric Tensors):

...there is a unique scalar product $g^1_1$ on $V \otimes V^*$ such that for $v_1 \otimes \alpha_1$ and $v_2 \otimes \alpha_2$ $\in V \otimes V^*$ we have $$ g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) \, g^*(\alpha_1, \alpha_2)$$

Here $g^*$ is a scalar product on $V^*$: $$g^*(\alpha, \beta) = g(\alpha^\sharp, \beta^\sharp)$$

What bothers me is that I do not understand why it couldn't be $$ g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) + g^*(\alpha_1, \alpha_2)$$

Is there a reason to choose the first definition? Particularly given how metric tensor is defined of products of Riemannian manifolds.

For $V \otimes V$ the book earlier defines scalar product as: $$ g \left(v_1 \otimes u_1), (v_2 \otimes u_2) \right) = g(v_1,v_2) \, g(u_1, u_2)$$

Answer 1

The map $$g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) + g^*(\alpha_1, \alpha_2)$$ isn't even well-defined: For all $\lambda \in \mathbb{R} - \{0\}$ we have $$v_1 \otimes \alpha_1 = (\lambda^{-1} v_1) \otimes (\lambda \alpha_1),$$ but $$g(\lambda^{-1} v_1,v_2) + g^*(\lambda \alpha_1, \alpha_2) = \lambda^{-1} g(v_1,v_2) + \lambda g^*(\alpha_1, \alpha_2),$$ which in general does not coincide with $$g(v_1,v_2) + g^*(\alpha_1, \alpha_2).$$

What you probably have in mind is this: Given scalar products $g, h$ respectively on vector spaces $\mathbb{V}, \mathbb{W}$ we get a natural scalar product on the direct sum $\mathbb{V} \oplus \mathbb{W}$ defined by $$((v_1, w_1), (v_2, w_2)) \mapsto g(v_1, v_2) + h(w_1, w_2)).$$

Answer 2

If you multiply $v_1$ by $\lambda$ and $\alpha_1$ by $1/\lambda$, you don't change their tensor product $v_1 \otimes \alpha_1$. Hence any formula that one proposes for the scalar product must also be left unchanged by such an operation, and this rules out your formula with "plus" instead of "times".

Answer 3

I think it might be interesting to consider why the scalar product is defined that way:

Recall that the tensor product $V\otimes W$ is constructed by taking $\mathrm{Free}(V\times W)$, the vector space freely generated by the elements in $V\times W$, and dividing out by the subspace $N_{V\otimes W}\subset \mathrm{Free}(V\times W)$ of all elements generated by $(v,w) + (v',w) - (v+v',w)$, and another two or three relations that you can easily find.

If you have any two maps $f\colon V\to V'$ and $g\colon W\to W'$, then it is certainly natural to consider the product map $(f,g)\colon V\times W \to V'\times W', (v,w)\mapsto (f(v),g(w))$. This extends to a linear map on

$$ (f,g)\colon \mathrm{Free}(V\times W) \to \mathrm{Free}(V'\times W') \;, $$

because it is already defined on the generators of $\mathrm{Free}(V\times W)$.

To show that $(f,g)$ descends to a well-defined map $$ f\otimes g\colon V\otimes W \to V'\otimes W' $$ it suffices to show that equivalence classes on the left are mapped to equivalence classes on the right, which boils down to verifying identities like

$$ (f,g) \bigl((v,w) + (v',w) - (v+v',w)\bigr) = (f,g) \bigl((v,w)\bigr) + (f,g)\bigl((v',w)\bigr) - (f,g)\bigl((v+v',w)\bigr) = (fv,gw) + (fv',gw) - (f(v+v'),gw) = (fv,gw) + (fv',gw) - (fv+fv',gw) \;.$$

Since $(f,g)\bigl(N_{V\otimes W}\bigr) \subset N_{V'\otimes W'}$, we have a well-defined $f\otimes g$; linearity is automatic.

I haven't done the exercise, but there is no doubt that the definition of the scalar product on a tensor space follows directly from the analogue of the construction above for bilinear maps on tensor products.