Start a new discussion

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeFeb 10th 2012

I found in the disambiguation page basis an entry basis of a vector space missing, and so I created one. I have then cross-linked this with relevant existing entries, such as basis theorem. Also created a brief entry orthogonal basis in order to get rid of grayish links is some entries.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeFeb 11th 2012

The last now also has orthonormal basis and a link to Gram–Schmidt process.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 11th 2012

Except that we now have internal inconsistency: an orthonormal or orthogonal basis is not a basis (of the algebraic kind) for an infinite-dimensional Hilbert space. I was going to enter a correction/warning, but I first wanted to ask: does the standard definition of inner product over a general field include nondegeneracy? I’m having trouble getting an answer from google (most sources are for the case over $\mathbb{R}$, $\mathbb{C}$), but I thought I had a dim memory of nondegeneracy from somewhere.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeFeb 11th 2012

Wikipedia’s inner product contains positive-definiteness. On the other hand, the entry indefinite inner product says it is an “indefinite inner product” and so contradicts that design choice.

I think this is a classical case where we should not try to get a globally consistent convention, but rather point out the different variants in each entry.

• CommentRowNumber5.
• CommentAuthorzskoda
• CommentTimeFeb 11th 2012
• (edited Feb 11th 2012)

Urs, Todd asked “over a general field”, while the wikipedia is explicitly saying “the field of scalars is…either $\mathbf{R}$ or $\mathbf{C}$”.

For me it is nondegenerate but I hear and sometimes say myself explicitly when introducing the data “nondegenerate inner product”. The positive definiteness is even more rigid and less spread requirement in general situation, I stumble often into indefinite inner products, even over $\mathbf{R}$ and $\mathbf{C}$.

One should also consult the convention in the theory of Hilbert modules.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeFeb 11th 2012

Well, even there, Wikipedia is inconsistent in itself, as I pointed out.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 11th 2012

I made some changes to orthogonal basis.

I don’t care what Wikipedia says in this instance; I do care what the nLab convention is to be for inner product space. Non-degenerate or not?

• CommentRowNumber8.
• CommentAuthorzskoda
• CommentTimeFeb 11th 2012

Isn’t this the red herring, white herring business in fact ?

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 11th 2012

Zoran, yes, and I actually linked to red herring principle in orthogonal basis.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeFeb 11th 2012

My initial inclination would be to require an inner product over a general field to be nondegenerate. I feel like nondegeneracy is important to a lot of how I think of “inner product spaces” behaving; if it weren’t nondegenerate I would be more inclined to call it “a space equipped with a quadratic form” or something.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeFeb 11th 2012
• (edited Feb 11th 2012)

I don’t care what Wikipedia says in this instance; I do care what the nLab convention is

I still think it is unwise to try to impose a convenion on a wiki if it is not adhered to in practice. Therefore I said:

I think this is a classical case where we should not try to get a globally consistent convention, but rather point out the different variants in each entry.

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 12th 2012

It actually wasn’t clear to me what exactly you were referring to in #4, Urs. It sounded like you were saying there is some inconsistency in Wikipedia (as you repeated in #6), which is irrelevant to my concerns here. But anyway, do you think it is impossible to come to agreement on nondegeneracy of inner product here, that it will be impossible to adhere to in practice? Why should we not try to get a globally consistent convention on this matter; is it really so difficult?

• CommentRowNumber13.
• CommentAuthorTobyBartels
• CommentTimeFeb 20th 2012
• (edited Feb 20th 2012)

I disagree that the red herring principle applies as it appears in the article. An orthonormal basis is a basis. However, when working with topological vector spaces, one uses the term ‘basis’ differently from the strictly algebraic sense (called ‘Hamel basis’ by analysts). Interestingly, no infinite-dimensional positive-definite inner-product space has an orthonormal basis in the algebraic sense. However, every Hilbert space (assuming the axiom of choice) has an orthonormal basis in the topological sense.

• CommentRowNumber14.
• CommentAuthorTobyBartels
• CommentTimeFeb 20th 2012

Thinking that I am certainly right, I rewrote much of Todd’s contribution. Todd, if I lost anything of importance, please put it back. And if you disagree with me more than that, then (edit how you like and) we should talk it out.

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 20th 2012
• (edited Feb 20th 2012)
Hi Toby. I see what you mean about red herring, and I'm fine with having it removed. However, I felt some clarifications were still in order, so I rewrote orthogonal basis one more time. Please let me know whether you agree with how it's now written.
• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 20th 2012
By the way, I think more clarifications might be in order at basis and at basis in functional analysis, but I'll have to report back on these later.
• CommentRowNumber17.
• CommentAuthorTobyBartels
• CommentTimeFeb 21st 2012

Looks good, Todd! And I think that you mean (instead of or in addition to basis) basis of a vector space.

• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 21st 2012
• (edited Feb 21st 2012)

Thanks, Toby. I think basis of a vector space looks okay to me. Elsewhere, I have the following quibbles:

• At basis, where it says “generating set of a free module”. That’s fine for the algebraic sense of basis; I’m not sure it’s fine for the functional analysis sense. For example, the elements of a set $X$ can be viewed as a linearly independent set in $l^3(X)$ whose span is dense in $l^3(X)$, but in which category is $l^3(X)$ viewed as a free module? (In the category of Banach spaces and norm-decreasing maps, $l^1(-)$ is left adjoint to the forgetful functor to $Set$.) I can think of more awkward circumlocutions, but I thought I should bring it up first.

• Over at basis in functional analysis, under topological basis – do we really mean to restrict to limits of sequences?

• CommentRowNumber19.
• CommentAuthorTobyBartels
• CommentTimeFeb 22nd 2012

In the first case, I think that I fixed it. In the second case, we must not, since the alternative definition has nothing sequential in it. So I fixed that too.

• CommentRowNumber20.
• CommentAuthorTim_Porter
• CommentTimeFeb 22nd 2012
• (edited Feb 22nd 2012)

This is just idle curiosity. I have not been following this thread, but wonder if there is a mention of the ideas around ’invariant basis number’ . I remember the book ‘Free rings and their relations’ by Cohn, which had a lot about that topic. I do not know if the ideas have ’survived’ into active mathematical topics for today.

• CommentRowNumber21.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 22nd 2012

@Toby: thanks!

@Tim: I think this is probably worth a mention in various spots around the nLab, such as one of the basis articles, and certainly at counterexamples in algebra (an example of a ring that fails IBN, if one isn’t not already there). I’ll look around in a bit.

• CommentRowNumber22.
• CommentAuthorTim_Porter
• CommentTimeFeb 22nd 2012
• (edited Feb 22nd 2012)

The quick fix is to link to wikipedia, but the point that not all rings have the IBN property is worth signalling. This MO question/answer is good: here

• CommentRowNumber23.
• CommentAuthorAndrew Stacey
• CommentTimeFeb 23rd 2012

I noticed something at basis in functional analysis that I’m not sure about (and I note that I wrote it in the original!). It is claimed that a topological basis has a dual basis. Is this true for a topological vector space without the assumption of local convexivity? One certainly needs Hausdorff: simply take $\mathbb{R}^2$ with the indiscrete topology. Then a single vector $v \in\mathbb{R}^2$ is a topological basis but applying the argument on that page to the family $\{v\}$ doesn’t work because we end up with a codimension 2 subspace, not codimension 1. I don’t know much about non-locally convex tvs, but certainly if we discard the Hahn-Banach theorem then we get the fun counter-example that $\ell^\infty/c_0$ has no linear functionals, and so a topological basis for $\ell^\infty$ would have severe difficulties in having a dual basis.

On orthogonal basis, I teach these as maximal orthogonal families, arguing that a basis in finite dimensions can be thought of as a maximal linearly independent set - basically because I don’t want to get into too much unnecessary detail on topologies.

• CommentRowNumber24.
• CommentAuthorTobyBartels
• CommentTimeFeb 23rd 2012

Actually, a singleton $B = \{v\}$ is not a topological basis of a vector space with the indiscrete topology, because $v$ is a limit of the sequence $(0,0,0,\ldots)$, all of whose elements are linear combinations of elements of $B \setminus \{v\}$. Rather, the empty set $B = \varnothing$ is a basis (so topologically, the dimension is $0$). And it has a dual basis, itself. I think that the existence of a dual basis works fine in the non-Hausdorff case, with these caveats: the argument presented in the article doesn’t directly apply, the dual basis is not given uniquely but only up to topological indistinguishability (so we may replace any element $v$ of it by any element of $\overline{\{v\}}$) and may not exist at all without the axiom of choice (but of course we’re already using Hahn–Banach). You prove this by passing to the Hausdorff quotient and applying the argument in the article.

Still, everything is simpler in the Hausdorff case.

I cannot help with the non-locally-convex case. The argument presented doesn’t apply there, and I can’t fix it either.