# Trying to Understand the Hodge Star

Discussion in 'Physics & Math' started by Reiku, Dec 7, 2011.

1. ### rpennerFully WiredRegistered Senior Member

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Along those lines, if you count from 0 to $2^N-1$ in binary, and group those numbers by how many 1's each binary representation has, you are illustrating your idea. If, in addition, you drawn lines between each pair of elements which differ by exactly one bit, then you have a graph of a N-cube.

Last edited: Dec 14, 2011

3. ### arfa branecall me arfValued Senior Member

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That diagram looks like an upside-down version of a certain Boolean lattice.
The nodes labelled 1,2,4,8 could be the four possible outcomes of tossing two coins.

The node labelled 0 is then the meet of the four nodes above it (which are below it in the diagram). 0011 is the join of 0001 and 0010, the two nodes below it.
You get 16 nodes because that's the size of the powerset of {0001,0010,0100,1000}.

5. ### James RJust this guy, you know?Staff Member

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Mister:

I have urged you on several occasions to read the site rules for your own good. One of those rules is that if you create a sock puppet to evade a temporary ban then BOTH the sock and the original identity will be PERMANENTLY banned.

In this instance, I will do you a favour and once again assume you are too stupid to inform yourself of the rules. So, your "Mister" identity" is temporarily banned for now. When that ban expires you may come back. If you EVER create another sock puppet, ALL your identities will be PERMANENTLY banned. And, since this is your second chance here, there will be no third chance.

In post #17 of this thread you wrote "The Hodge star is a linear functional for differentials. It changes the k-forms to n+k forms."

So are you saying the Hodge star doesn't involve functionals now?

I don't think I said you had any cut and pastes about functionals.

Because of something AlphaNumeric wrote. I seem to recall something about the Hodge star having to do with functionals. And, to quote AlphaNumeric "Differential forms, specifically 1-forms, are functionals on the space of vectors."

If the Hodge star has to do with functionals, then you need to know what a functional is to understand what the Hodge star is, no?

I didn't say that. The topic has been discussed here before, if I recall correctly. What I said was that I don't properly understand what a Hodge star is or does.

But it appears you lack the mathematical apparatus to understand any work that would involve the Hodge dual.

It wouldn't take much of an adjustment in your approach to fit right in at sciforums. But this pretence that you're an accomplished physicist just repeatedly leads you to tell lies and make stuff up. That in itself is not a bit issue, except that it could mislead other uneducated readers. And that is a big problem. We really can't have you pretending to be an authority on physics and maths when so clearly you are not.

Ok, let me know when you get back what you want to do about this. I can ban you if that's what you want. You can't say you weren't given a fair go.

7. ### arfa branecall me arfValued Senior Member

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Slight correction: I should have had: 0 is the meet of any two of the four nodes above it. Again, the diagram has the direction (but not the logic) inverted.

The relation: "is different by one bit", can't be reflexive or transitive, can it?

8. ### AlphaNumericFully ionizedRegistered Senior Member

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Do the parity flip $0 \leftrightarrow 1$ and you turn it upside down so it's equivalent to the usual representation.

9. ### funkstarratsknufValued Senior Member

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No, so it's not actually a lattice, because that relation doesn't define a partial order.

10. ### arfa branecall me arfValued Senior Member

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That relation doesn't, but are you sure there isn't a partial order?

There are 5 partitions, aren't there? So what does that?

Edit:
What does it is whatever 'counts' the 1s in each 4-bit string, right?
In the case of {0001, 0010, 0100, 1000}, each distinct number represents a Boolean term, like (p and q), say. So p has to be a 4-bit string such that (p and q) is one of the numbers in this set, with one 1.

Last edited: Dec 16, 2011
11. ### funkstarratsknufValued Senior Member

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No, that doesn't work either: First, it's not anti-symmetric. $|0011|_1 \leq |1100|_1$ and $|1100|_1 \leq |0011|_1$ but $0011 \neq 1100$. Second, it relates element in each "level" of the diagram, but the diagram has no edges there.

One relation which actually works to define the partial order is $x~R~y$ iff $x \leq y$ and $|x|_1 \leq |y|_1$. (Again, upside down of the diagram, but no matter.)

12. ### arfa branecall me arfValued Senior Member

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So, the meet or join relation doesn't make it a lattice either?

It's just that I have an article about this particular structure that says it's a Boolean lattice, so I'm interested in why you think it isn't.

13. ### funkstarratsknufValued Senior Member

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I don't understand what you mean by this. There's only one relation.
Oh, it is a lattice, but e.g by the relation I specified.

My previous comments were to the effect that the relations you suggested didn't work, because they didn't actually reflect the order in the diagram, nothing more.

14. ### arfa branecall me arfValued Senior Member

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But meet and join are dual relations, aren't they?

But you said "it's not actually a lattice". What you meant was, "with that relation you don't have a lattice".

So that's been cleared up then.

When I said "whatever 'counts' the 1s in each string", I meant the numbers are obviously partitioned into strings with zero, one, two, three, or four 1s. I don't understand yet why you said it doesn't work.

Last edited: Dec 16, 2011
15. ### funkstarratsknufValued Senior Member

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I still don't know what you mean by this. The underlying relation of a lattice is a partial order.
I gave you a concrete counterexample? "0011" has the same number of 1s as "1100". "1100" has the same number of 1s as "0011". Lattices are posets so by definition the order has to be anti-symmetric. That means that the order R, whatever it is, has to respect that $x~R~y$ and $y~R~x$ implies $x=y$. But "0011" and "1100" are not the same, so "has the same number of 1s" is not a partial order, and thus by definition cannot define a lattice on strings.

Also, there are edges (meaning that they are comparable in the order) between, say, "0001" and "0011", which manifestly don't have the same number of 1s. There are no direct edges between, say "0011" and "1100", which do. Finally, note that you cannot get from "0011" to "1100" without going both up and down in the order, meaning that these two elements are incomparable.

Unless you're suggesting an order of sets of strings with the same number of ones? (Though that is clearly not the structure of the diagram)

16. ### arfa branecall me arfValued Senior Member

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Whatever partitions the strings into 5 subsets, each subset is an antichain. So the maximum chain must be length 5. "Has the same number of 1s" may not be a partial order relation, but it does exist as a relation.

That's really all I was saying. If you want Boolean 'counting', I guess you need to use what passes for arithmetic in Boolean logic, which is (combinations of) and, or, and not.
This implies that each string corresponds to a logical statement about the two variables p and q (which are like two coins, i.e. are true or false).

p.s. on p39 of "Introduction to Lattices and Order" 2nd ed by Davey and Priestley, is the following:

Last edited: Dec 17, 2011
17. ### funkstarratsknufValued Senior Member

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OK, then I just don't understand why you were saying it. It's not the order of the lattice, and it's not a partial order on 0-1 strings. What was your point in bringing it up, then?
I still don't understand what your point is with this.

18. ### arfa branecall me arfValued Senior Member

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I was only repeating what rpenner said:

19. ### James RJust this guy, you know?Staff Member

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Is this thread still about the Hodge star, or should we close it and start a new thread?

20. ### funkstarratsknufValued Senior Member

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Aside from the combinatorics, I don't think there's any deeper connection - I was merely commenting on the lattice theory stuff. Frankly, before this thread appeared I wasn't even aware of the Hodge star.

21. ### ReikuBannedBanned

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11,238
So James and Funkstar weren't aware of a hogde star. But god forbid me actually making a thread on it. I mean, it doesn't matter if I get something wrong, the whole point is to humiliate me anyway.

22. ### QuarkHeadRemedial Math StudentValued Senior Member

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That admission by JamesR and funkstar is an instance of what is called humility. Pound to a penny if either were sufficiently motivated they would easily understand the subject because they have the background knowledge

My guess - well it is rather more than that - is that you do not have that background knowledge, and are asking questions whose answers you are not equipped to understand.

The attempts to what you call "humiliate" you are merely attempts to persuade to adopt the same sort of humility.

Remember - ignorance is not a sin, but can only be cured (mixed metaphor!) by long and hard study.

Start today

23. ### rpennerFully WiredRegistered Senior Member

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From posts #95, #96.
That is a list of n objects, presumably vectors.
The definition of a vector space is more general.

Pick a field, $F$, like $\mathbb{R}$ or $\mathbb{C}$.
Then a vector space is a mathematical Abelian group, written with operator $+_v$, and a scalar multiplication law, written as $\cdot_s$ that is compatible in a certain way with the field. Such that if $f_1, f_2, f_3 \in F$ and $v_1,v_2,v_3 \in V$ we have:
Because $F$ is a Abelian group with respect to addition:
$f_1 \, + \, ( f_2 \, + \, f_3 ) \; = \; ( f_1 \, + \, f_2 ) \, + \, f_3$
$f_1 \, + \, f_2 \; = \; \, f_2 + \, f_1$
$\exists 0 \in F \quad 0 \, + \, f_1 \; = \; f_1 \, + \, 0 \; = \; f_1$
$\exists -f_1 \in F \quad -f_1 \, + \, f_1 \; = \; f_1 \, + \, -f_1 \; = \; 0$

Because $F$ (with the exception of 0) is a Abelian group with respect to multiplication:
$f_1 \, \cdot \, ( f_2 \, \cdot \, f_3 ) \; = \; ( f_1 \, \cdot \, f_2 ) \, \cdot \, f_3$
$f_1 \, \cdot \, f_2 \; = \; f_2 \, \cdot \, f_1$
$\exists 1 \in F \quad 1 \, \cdot \, f_1 \; = \; f_1 \, \cdot \, 1 \; = \; f_1$
$f_1 \ne 0 \quad \Rightarrow \quad \exists f_1^{-1} \in F \quad f_1^{-1} \, \cdot \, f_1 \; = \; \, f_1 \cdot \, f_1^{-1} \; = \; 1$

Because $V$ is a Abelian group with respect to vector addition:
$v_1 \, +_v \, ( v_2 \, +_v \, v_3 ) \; = \; ( v_1 \, +_v \, v_2 ) \, +_v \, v_3$
$v_1 \, +_v \, v_2 \; = \; v_2 \, +_v \, v_1$
$\exists 0_v \in V \quad 0_v \, +_v \, v_1 \; = \; v_1 \, +_v \, 0_v \; = \; v_1$
$\exists -v_1 \in V \quad -v_1 \, +_v \, v_1 \; = \; v_1 \, +_v \, -v_1 \; = \; 0_v$

Because mathematicians found it a pain to exclude the special ( but trivial ) 1-element field: $1 \ne 0$ This still leaves 2 big open questions about the field: If you add 1's many times, do you ever wind up at zero? For $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ the answer is no. Another open question is if there are any elements of the field which you can't get to starting with 0, 1 addition, multiplication and their inverses .For $\mathbb{Q}$ the answer is no. For $\mathbb{R}$ the answer is yes, with some examples being $\sqrt{2}, \pi, e, \dots$

Because $F$ is a field, and not just two groups with the same elements, it has a distributive law:
$f_1 \, \cdot \, ( f_2 \, + \, f_3 ) \; = \; f_1 \, \cdot \, f_2 \; + \; f_1 \, \cdot \, f_3$

Finally, because scalar multiplication relates F to V, we have the operation of scalar multiplication which produces vectors.
$1 \, \cdot_s \, v_1 \; = \; v_1$
$f_1 \, \cdot_s \, ( f_2 \, \cdot_s \, \v_1 ) \; = \; ( f_1 \, \cdot \, f_2 ) \, \cdot_s \, v_1$
$( f_1 \, + \, f_2 ) \, \cdot_s \, v_1 \; = \; f_1 \, \cdot_s \, v_1 \; +_v \; f_2 \, \cdot_s \, v_1$
$f_1 \, \cdot_s \, ( v_1 \, +_v \, v_2 ) \; = \; f_1 \, \cdot_s \, v_1 \; +_v \; f_1 \, \cdot_s \, v_2$
From these axioms, you can prove additional neat relationships. Usually we know what objects we are working with and drop the subscripts on the operators and zeros.

For example if you have a set of vectors: $\left{ v_1, \dots v_m \right}$ you can ask the question, is there an indexed set of scalars (not all 0) in the field such that $\sum_{k=1}^m \quad f_k \, \cdot_s \, v_k \; = \; 0_v$ ? If no set of scalars allows this, then the set of vectors is said to be "linearly independent".

And with that preamble, the definition of an n-dimensional vector space is a vector space which supports at most linearly independent sets of size n.

Since an n-dimensional vector space does not support a linearly independent sets of vectors of size n+1, then it follows that given a linearly independent set of size n called B, and any vector v then there exist an indexed set of scalars such that $\sum_{k=1}^n \quad a_k \, \cdot_s \, b_k \; = \; v$ and such "linearly independent" set of n vectors forms a basis for $V_n$.

Yay!

Specifically, the dimension of $\Lambda^p(V_n)$ is ${n \choose p}$.

What was meant was for you to write the dimension of $\Lambda^p(V_n)$ in terms of factorials.

What was meant was for you to write the dimension of $\Lambda^{n-p}(V_n)$ in terms of factorials.

Last edited: Dec 19, 2011