Trying to Understand the Hodge Star

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

1. Fudge MuffinFudge MuffinRegistered Senior Member

Messages:
148
I think he actually might be Stewie.

3. ReikuBannedBanned

Messages:
11,238
Am I talking to a cheesecake or something?

I replied to you the first time, stating that I did not mean functionals, but rather function.

Shall I repeat it for a fourth time, maybe fifth?

5. ReikuBannedBanned

Messages:
11,238
Really nice post. Won't have time to go over it today. I appreciate you have actually spent time on this!

7. ReikuBannedBanned

Messages:
11,238
I'm glad your set theory on me is on it's limit. Not my strongest point, but I can read that off. Maybe what confused me about quarks post is he never took time to explain his notation. One problem I have is what is $\Lambda$ exactly? I take it $V$ is a vector, is that what lambda is?

I can attach the resemblence of it to the binomial coefficient. How it computes is something I am not sure of. Can someone explain this, in nice terms please.

8. QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,543
Now where does set theory enter into rpenner's post?

First, no, V is not a vector, it is a vector space

Second, as I said, the notation is arbitrary. But the $\Lambda$ on its own, in this context at least, makes no sense.

So I was trying to be gentle with you dear boy, but if you want the full Monty......

Given the vector space $V_n$ of dimension $n$, then the space $\Lambda^p(V_n)$ is called the p-th exterior power of this space.

Any wiser?

9. ReikuBannedBanned

Messages:
11,238
Well I see some ''exists'' symbols in there coupled with ''in'' symbols, purely set theory classical algebra

$\exists 0 \in F$

If that was gentle, you will need to be gentler. don't think this is a reflection on your teaching, but me in general. I need books which explain all the terminology correctly, with definition, so I am not confused. I have an illness which tends to mistake things for other things.

10. QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,543
Let me be more gentle, then. First set theory itself is NOT an algebra. Exercise: Why?

The fact that set theoretic symbols are widely used in all branches of mathematics has to do with the fact that all these branches (group theory, linear algebra, topology, manifolds,.....) ALL start with a set and then proceed to impose an algebraic structure on it.

Exercise related to the previous: Define "an algebraic structure"

11. AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,697
Set theory and linear algebra are not synonymous.

It's standard notation in differential forms and tensors, the basic language of the Hodge star.

Do you understand tensors and differential forms? This isn't a rhetorical question.

12. rpennerFully WiredRegistered Senior Member

Messages:
4,833
From Posts #120, #124, #125
It doesn't because at no point did I identify anything as a set. These axioms hold even if $F$ or $V$ is a proper class or an arbitrary container metaphor.

So you don't need set theory since you aren't making infinite collections of objects -- you just need the symbols of F and V to limit the universe of objects you discuss to the field and the vector space in question.

As an exercise, on can describe an Abelian group like these without the existential qualifiers. But you need both a binary operator: $+_g$ and an unary operator: $-_g$.

$((g_1 \, +_g \, g_2) \, + \, g_3) \, +_g \quad -_g \, ( g_1 \, +_g \, g_3) \; = \; g_2$
Or colloquially $((a + b)+c)-(a + c) = b$

Ref: William McCune in 1992.

But then it takes cleverness and a bit of work to recover the standard theorems and axioms of group theory.
Example: $x + x = x \Rightarrow x - x = x$ follows from $((a + b)+c)-(a + c) = b$ without assuming any other properties of the operators + and -. Likewise $x + x = x \quad \textrm{and}\quad y = -x \Rightarrow x = y$. It's somewhat cumbersome, but without set theory you can prove that a particular Abelian group of a certain size is symmetric from the McCune axiom.
With small finite groups represented as tokens in a bag, it's just a matter of pure logic to enumerate all possibilities. Like sudoku -- with more bookeeping.

Well this thread prompted me to know that much before replying. There's a reason why math and physics educators spoon-feed predigested math and simpler than state-of-the-art physics models at the beginning of education and progress as time goes on. It's partly because learning everything is beyond at least 99.999% of humanity and it's partly an expression of the finite amount of time and resources humans and human organizations have to allocate to education.

I was taught (Maxwell's) electromagnetism at an undergraduate level without being exposed to group theory, tensors or the functional calculus. But I absolutely had to have mastered the tools of multivariable calculus and have a practiced geometric intuition to solve the problems assigned to me. This mandatory foundation is described in the course catalog as "prerequisite" as in required topics to have been mastered previously so you won't cause the lecturer to write his dean and ask "what's the matter with kids today?" Since going to university cost a good deal more than my entire discretionary budget at the time, and participation on this forum is free, why would anyone think that there would be less onerous prerequisites to mastering a mathematical topic online?

13. ReikuBannedBanned

Messages:
11,238
Oh, I am so glad the rocket scientist came in and stated thatt set theory in standard notation are the basic language of a Hodge star.

Now... will his highness please tell me why I would be privy to such things, when my post clearly highlights I know near to nothing about a hodge dual?

Lol

I do love it when people state the obvious!

14. ReikuBannedBanned

Messages:
11,238
No you never, but I am under the impression you are recreating a rule here to say that if set theory applications are being met but not mentioned, then it doesn't count.

15. ReikuBannedBanned

Messages:
11,238
Actually, set theory is an analog of algebra.

You can explain for instance, Bells Inequality in a series of set theory. In fact, it is the perfect analog of such a statement as his.

16. ReikuBannedBanned

Messages:
11,238
Is this a trick question? If I had the technology, I'd easily show that set theory is a perfect analog to Bells Inequality, which is an algebra.

Secondly, using set theory to describe a succession of equations does not defeat the fact the equations Rpenner used was of a set theory analog. Whether it was a set theory directly, is arguable. Anything which contains ''in'' and ''exists'' symbols are quite clearly from the set theory catagory.

17. AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,697
I didn't say that. I said that $\Lambda$ is used as a particular notion in differential forms. It's almost in the definition.

So if you admit you don't know terminology used in the definition of differential forms why are you asking about things which alter differential forms? You have just proven you start topics where you don't even understand the definitions of things you talk about. Thus rendering this thread as little more than an attempt to talk about things you don't understand.

You just demonstrated you don't pay attention. Again. Notice what QH said, he said "AN algebra". He referred to a specific algebra. In mathematics before university 'algebra' means doing stuff with x and polynomials and a bit of calculus. All the stuff where you use variables and not just specific numbers. In proper mathematics (seriously, you wouldn't believe how far much of high school mathematics is from proper mathematics!) an algebra is a specific type of concept, like group or set or vector space.

An algebra is a vector space with a binary product $\ast$ defined on it. For example, you can turn $\mathbb{R}^{N}$, which is a vector space, into an algebra by defining something which converts a pair of vectors into a vector. For N=3 an example is the cross product, ie $a \ast b \equiv a \times b$. That is an algebra.

Again, this is basic stuff you cover well before you get to differential forms. You've skipped to the 'cool stuff' (algebras are cool though but that's just me) and in doing so you've missed ESSENTIAL ground work. It's what so easily gives your deception away.

Does that mean anything with 'field' in is to do with agriculture?

18. wlminexBannedBanned

Messages:
1,587
a farmer is . . . "a man outstanding in his field" . . . as are you AN!

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

Messages:
31,079
wlminex:

Please stop posting inane personal commentary in the science forums. Keep it for the Fringe areas. In fact, probably better if you restrict your activities to those forums in general. You don't seem to contribute anything useful in scientific discussions.

20. QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,543
So having strutted the high-and-mighty, let me ask this blushingly naive question:

Is it a requirement that an algebra have a derivation (i.e. satisfy the Law of Leibniz), or does it follow naturally from more primitive axioms?

Are there any algebras that do not have a derivation?

21. rpennerFully WiredRegistered Senior Member

Messages:
4,833
We have a field, F, and a vector space V.
And we have operators $+$, $\cdot$, $+_v$, $\cdot_s$ and the axioms above.
To that we add a vector product (returning vector) $\cdot_v$ and a distributive law and a law of compatibility with scalar multiplication.
$(v_1 \; +_v \; v_2)\; \cdot_v \; (v_3 \; +_v \; v_4) = v_1 \, \cdot_v \, v_3 \; +_v \; v_1 \, \cdot_v \, v_4 \; +_v \; v_2 \, \cdot_v \, v_3 \; +_v \; v_2 \, \cdot_v \, v_4$
$f_1 \, \cdot_s \, ( v_1 \, \cdot_v \, v_2 ) = ( f_1 \, \cdot_s \, v_1 ) \, \cdot_v \, v_2 = v_1 \, \cdot_v \, ( f_1 \, \cdot_s \, v_2 )$
Finally, a Derivation is an operator, D, that satisfies
$D ( v_1 \, +_v \, v_2 ) = D ( v_1 ) \; +_v \; D( v_2 )$
$D ( f_1 \, \cdot_s \, v_1 ) = f_1 \, \cdot_s D ( v_1 )$
$D ( v_1 \, \cdot_v \, v_2 ) = D ( v_1 ) \, \cdot_v \, v_2 \; +_v \; v_1 \, \cdot_v \, D( v_2 )$
So for any field and k an element of that field, the operator: $k \cdot_s I$ seems to satisfy all axioms but the last (the law of Leibniz).
And for all fields, the operator $0 \cdot_s I$ satisfies the law trivially.
This is possibly the only type of derivation for a 1-dimensional algebra.

22. AlphaNumericFully ionizedRegistered Senior Member

Messages:
6,697
In the sense of an algebra over a ring there's no need for a derivative at all. A derivative needs the Leibnitz rule (it's almost a definition) but a bilinear map doesn't need a derivative to be bilinear. The cross product on $\mathbb{R}^{3}$ gives an algebra without the need for a derivative.

Unless we're talking about different algebras?

23. funkstarratsknufValued Senior Member

Messages:
1,390
Sign me up for this. When I hear "this is an algebra" I think "this has some algebraic structure of some sort" without any particular structure in mind...