Question about Hodge duality and inner products

I'm looking at this definition on the wiki:

http://en.wikipedia.org/wiki/Hodge_dual

It says

"In mathematics, the Hodge star operator or Hodge dual is a
significant linear map introduced in general by W. V. D. Hodge. It is
defined on the exterior algebra of a finite-dimensional oriented inner
product space."

Keenan has drummed into my head that most of regular temperament
theory works without any sort of inner product being defined, until
you get into things that use the TE norm or what have you. So is the
above correct? What is the inner product that is needed to say that
|-4 4 -1> and <<4 4 1|| are Hodge dual to one another?

-Mike

Mike Battaglia <battaglia01@gmail.com> wrote:

> Keenan has drummed into my head that most of regular
> temperament theory works without any sort of inner
> product being defined, until you get into things that use
> the TE norm or what have you. So is the above correct?
> What is the inner product that is needed to say that |-4
> 4 -1> and <<4 4 1|| are Hodge dual to one another?

That's a Euclidean norm, isn't it?

Graham

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> Keenan has drummed into my head that most of regular temperament
> theory works without any sort of inner product being defined, until
> you get into things that use the TE norm or what have you. So is the
> above correct? What is the inner product that is needed to say that
> |-4 4 -1> and <<4 4 1|| are Hodge dual to one another?

You need the inner product, or at least a volume form, to decide that e2^e3 and e5 are in a dual relation. By the way, you want |-4 4 -1> and <<1 4 4||. I was going to refer you to the Xenwiki article on this topic, but apparently I never wrote one, so I guess I should.

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> Keenan has drummed into my head that most of regular temperament
> theory works without any sort of inner product being defined, until
> you get into things that use the TE norm or what have you. So is the
> above correct? What is the inner product that is needed to say that
> |-4 4 -1> and <<4 4 1|| are Hodge dual to one another?

I've been thinking about your Hodge question, and I can't think of where we would really need the Hodge dual rather than just the complement. How about a Xenwiki article on that?

You might use this dual as a step toward calculating the inner product. Or you might calculate the inner product a different way. I don't think you even need the special case of the complement.

Graham

------Original message------
From: genewardsmith <genewardsmith@sbcglobal.net>
To: <tuning-math@yahoogroups.com>
Date: Saturday, December 31, 2011 9:57:18 AM GMT-0000
Subject: [tuning-math] Re: Question about Hodge duality and inner products

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> Keenan has drummed into my head that most of regular temperament
> theory works without any sort of inner product being defined, until
> you get into things that use the TE norm or what have you. So is the
> above correct? What is the inner product that is needed to say that
> |-4 4 -1> and <<4 4 1|| are Hodge dual to one another?

I've been thinking about your Hodge question, and I can't think of where we would really need the Hodge dual rather than just the complement. How about a Xenwiki article on that?

------------------------------------

Yahoo! Groups Links

On Dec 31, 2011, at 4:57 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> Keenan has drummed into my head that most of regular temperament
> theory works without any sort of inner product being defined, until
> you get into things that use the TE norm or what have you. So is the
> above correct? What is the inner product that is needed to say that
> |-4 4 -1> and <<4 4 1|| are Hodge dual to one another?

I've been thinking about your Hodge question, and I can't think of where we
would really need the Hodge dual rather than just the complement. How about
a Xenwiki article on that?

I thought the "complement" was the Hodge dual? That's what you told me and
Keenan on XA chat. You mentioned that "complement" was just the name some
textbook you were all using gave to Hodge duality, and if the wiki article
on Hodge dual was up at the time you'd have just shown everyone that.

-Mike

.

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I thought the "complement" was the Hodge dual? That's what you told me and
> Keenan on XA chat. You mentioned that "complement" was just the name some
> textbook you were all using gave to Hodge duality, and if the wiki article
> on Hodge dual was up at the time you'd have just shown everyone that.

Numerically, if you want the "answer", the actual numbers, that's it. But you are raising the issue of mathematical definitions, where the Hodge dual maps k-multivals to (n-k)-multivals. We don't need to do that. All we need is that a k-multival corresponds to an (n-k)-multimonzo, which removes your concern over the inner product being dragged into it.

I'd like to know a good go-to place on the Web for this stuff. The book I told you about was Grassmann Algebra, by Browne, and I recall not liking some things about it, but I'll take another look.

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I thought the "complement" was the Hodge dual? That's what you told me and
> Keenan on XA chat. You mentioned that "complement" was just the name some
> textbook you were all using gave to Hodge duality, and if the wiki article
> on Hodge dual was up at the time you'd have just shown everyone that.

I looked at Browne again, and I still don't like it. I'll see if I can sort this out.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> I looked at Browne again, and I still don't like it. I'll see if I can sort this out.

I tried to upload Browne, but without success; I've got another book in the new "books" folder on linear algebra from an exterior algebra point of view.

--- In tuning-math@yahoogroups.com, "gbreed@..." <gbreed@...> wrote:
>
> You might use this dual as a step toward calculating the inner product. Or you might calculate the inner product a different way. I don't think you even need the special case of the complement.

Grassmann introduced the word "complement" as something which can be seen as a special case of the Hodge dual, but since I don't know a word for the concept we really want (which the Hodge dual article on Wikipedia, for example, discusses but never names) I was hoping to use if for that.

Where was this mentioned on the Wiki? I don't see any canonical way to
identify vectors with 2-covectors in a 3-dimensional space, or any
generalization of this at all. The word "covector" doesn't even appear on
the page.

Is this particular duality something that has representation in the
mathematical literature? It's obviously one of the most useful concepts
ever. If the answer to this question is "no," it seems obvious to me that
<<1 4 4|| is the Smith dual to |-4 4 -1>.

-Mike

On Sat, Dec 31, 2011 at 11:58 AM, genewardsmith <genewardsmith@sbcglobal.net
> wrote:

> **
>
>
>
>
> --- In tuning-math@yahoogroups.com, "gbreed@..." <gbreed@...> wrote:
> >
> > You might use this dual as a step toward calculating the inner product.
> Or you might calculate the inner product a different way. I don't think you
> even need the special case of the complement.
>
> Grassmann introduced the word "complement" as something which can be seen
> as a special case of the Hodge dual, but since I don't know a word for the
> concept we really want (which the Hodge dual article on Wikipedia, for
> example, discusses but never names) I was hoping to use if for that.
>
>
>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Where was this mentioned on the Wiki? I don't see any canonical way to
> identify vectors with 2-covectors in a 3-dimensional space, or any
> generalization of this at all.

Sorry, I was thinking of the Hodge dual section of the Exterior algebra article, where it discusses the canonical isomorphism between the kth exterior power of the dual space and the (n-k)th exterior power of the original space.

On Sat, Dec 31, 2011 at 12:28 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> >
> > Where was this mentioned on the Wiki? I don't see any canonical way to
> > identify vectors with 2-covectors in a 3-dimensional space, or any
> > generalization of this at all.
>
> Sorry, I was thinking of the Hodge dual section of the Exterior algebra article, where it discusses the canonical isomorphism between the kth exterior power of the dual space and the (n-k)th exterior power of the original space.

OK, I see

http://en.wikipedia.org/wiki/Exterior_algebra#Hodge_duality

So what exactly is the duality? It looks like it says that the
canonical isomorphism associates a, which is a member of the kth
exterior power of V*, with i_a * sigma, which is an element of the
(n-k)th exterior power of the original space.

What is i? What is sigma? How does this definition, which as far as I
can tell has two undefined variables, end up turning into your thing
in which the coefficients flip and the sign of some of the elements
changes?

-Mike

This only works in three dimensional space. It ties in with the cross product of vector algebra.

Graham

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: <tuning-math@yahoogroups.com>
Date: Saturday, December 31, 2011 12:12:45 PM GMT-0500
Subject: Re: [tuning-math] Re: Question about Hodge duality and inner products

Where was this mentioned on the Wiki? I don't see any canonical way to
identify vectors with 2-covectors in a 3-dimensional space, or any
generalization of this at all. The word "covector" doesn't even appear on
the page.

Is this particular duality something that has representation in the
mathematical literature? It's obviously one of the most useful concepts
ever. If the answer to this question is "no," it seems obvious to me that
<<1 4 4|| is the Smith dual to |-4 4 -1>.

-Mike

On Sat, Dec 31, 2011 at 11:58 AM, genewardsmith <genewardsmith@sbcglobal.net
> wrote:

> **
>
>
>
>
> --- In tuning-math@yahoogroups.com, "gbreed@..." <gbreed@...> wrote:
> >
> > You might use this dual as a step toward calculating the inner product.
> Or you might calculate the inner product a different way. I don't think you
> even need the special case of the complement.
>
> Grassmann introduced the word "complement" as something which can be seen
> as a special case of the Hodge dual, but since I don't know a word for the
> concept we really want (which the Hodge dual article on Wikipedia, for
> example, discusses but never names) I was hoping to use if for that.
>
>
>

On Sat, Dec 31, 2011 at 12:39 PM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> This only works in three dimensional space. It ties in with the cross product of vector algebra.
>
> Graham

What only works in three dimensional space? The same basic concept
should hold in 4d space as well, where it's now 3-covectors you're
associating with vectors.

This guy talks a lot about how wedge products are advantageous to
cross products: http://www.av8n.com/physics/clifford-intro.htm

-Mike

Sorry, yes. The complement will always give you a covector with the same number of elements as the original vector. But you can also choose the basis so that the vector and covector look the same. Then the inner product becomes a dot product, as expected for Euclidean space.

Graham

------Original message------
From: Mike Battaglia <battaglia01@gmail.com>
To: <tuning-math@yahoogroups.com>
Date: Saturday, December 31, 2011 4:01:28 PM GMT-0500
Subject: Re: Re: [tuning-math] Re: Question about Hodge duality and inner products

On Sat, Dec 31, 2011 at 12:39 PM, gbreed@gmail.com <gbreed@gmail.com> wrote:
>
> This only works in three dimensional space. It ties in with the cross product of vector algebra.
>
> Graham

What only works in three dimensional space? The same basic concept
should hold in 4d space as well, where it's now 3-covectors you're
associating with vectors.

This guy talks a lot about how wedge products are advantageous to
cross products: http://www.av8n.com/physics/clifford-intro.htm

-Mike

------------------------------------

Yahoo! Groups Links

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> So what exactly is the duality? I

That's what I'd explain if I wrote an article, which I still haven't found a name for.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
>
> > So what exactly is the duality? I
>
> That's what I'd explain if I wrote an article, which I still haven't found a name for.

Let me pose a question: should I call it

(1) Dual
(2) Complement
(3) Something else

I'll say #1.

-Mike

On Sat, Dec 31, 2011 at 11:55 PM, genewardsmith <genewardsmith@sbcglobal.net
> wrote:

> **
>
>
>
>
> --- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...>
> wrote:
> >
> >
> >
> > --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
> >
> > > So what exactly is the duality? I
> >
> > That's what I'd explain if I wrote an article, which I still haven't
> found a name for.
>
> Let me pose a question: should I call it
>
> (1) Dual
> (2) Complement
> (3) Something else
>
>
>