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Stream: learning: questions

Topic: Inner products

Eric Forgy (Dec 04 2020 at 18:46):

I'm working with directed graphs and differential graded algebras (and maybe a functor between then) and have a question about terminology. I think the question is best expressed by analogy with differential forms on manifolds.

If $\alpha,\beta$ are $p$-forms, we have

$\langle\alpha,\beta\rangle = \int_{\mathcal{M}} \alpha\wedge\star\beta = \int_{\mathcal{M}} (\alpha,\beta) \text{vol}.$

Both $\langle\cdot,\cdot\rangle$ and $(\cdot,\cdot)$ satisfy the conditions of an inner product EXCEPT the codomain.

$\langle\cdot,\cdot\rangle$ results in an element of the underlying field.

$(\cdot,\cdot)$ results in a 0-form.

Do we call both of these inner products?

Matteo Capucci (he/him) (Dec 04 2020 at 18:49):

I think it depends on the linearity of $(\cdot, \cdot)$. What I mean is that maybe it is bi/sesquilinear wrt to 0-forms and thus you can consider it an inner products with respect to that 'field' (actually, 'ring', but does it make a difference in the definition of inner products? I don't think so)

Eric Forgy (Dec 04 2020 at 18:54):

Thanks Matteo. Is there a term for "inner product" of $A$-bimodules if $A$ is a cummtative ring?

Eric Forgy (Dec 04 2020 at 18:55):

Or is calling that "inner product" fine?

Eric Forgy (Dec 04 2020 at 18:58):

Ah, the nLab defines inner product over rings: https://ncatlab.org/nlab/show/inner+product+space

Eric Forgy (Dec 04 2020 at 18:58):

Good enough for me :blush:

John Baez (Dec 04 2020 at 19:40):

Eric Forgy said:

I'm working with directed graphs and differential graded algebras (and maybe a functor between then) and have a question about terminology. I think the question is best expressed by analogy with differential forms on manifolds.

If $\alpha,\beta$ are $p$-forms, we have

$\langle\alpha,\beta\rangle = \int_{\mathcal{M}} \alpha\wedge\star\beta = \int_{\mathcal{M}} (\alpha,\beta) \text{vol}.$

Both $\langle\cdot,\cdot\rangle$ and $(\cdot,\cdot)$ satisfy the conditions of an inner product EXCEPT the codomain.

$\langle\cdot,\cdot\rangle$ results in an element of the underlying field.

$(\cdot,\cdot)$ results in a 0-form.

Do we call both of these inner products?

With differential forms we'd call the field-valued inner product the "inner product" and the 0-form-valued inner product the "pointwise inner product".

Eric Forgy (Dec 04 2020 at 19:43):

That works for me in the discrete setting too. Thanks John :blush: