Can anyone familiar with Levi-Civita point...

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Can anyone familiar with Levi-Civita point me in the right direction with this? I must be messing something trivial up or forgetting some commutation.

Also, can anyone show this simply with differential forms? I don't know much about them except that they exist.

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delta(ij) . a(j) = a(i), not a(j)

this might fix it

sorry if you're doing some weird index switching and I'm an idiot

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Well for one the derivative operator doesn't commute with B.

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>>7789330

Yeah. I mean that doesn't really make much sense, "the derivative of nothing" in a normal algebraic equation.

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>>7789322

Fuck you OP I wanted to go to bed now I'm stuck on this shit.

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Should be about right.

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>>7789263

>can anyone show this simply with differential forms?

I can't using forms, but I can (to an extent) with something similar (and better imo) called Geometric Algebra, I'll only do the first bit now (because I'm tired) but if this thread is still active in the morning I might do the second part. So first I'm going to throw some identities at you. [eqn] \text{ Inner and outer products are dual}: ~~~\left ( A \cdot B \right )^{*} = A \wedge (B)^{*} [/eqn] Next [eqn] \text { The outer product and cross product are dual } ~~~ A \times B = \left ( A \wedge B \right ) ^{*} ~~~ \text {Further} ~~~ ( A \times B ) ^{*} = (A \wedge B ) ^{**} = - A \wedge B [/eqn] Finally [eqn] \text { The dual of a multivector}: ~~~ A^{*} = A \mathbb{I} ^{-1} ~~~ \text{ Where } \mathbb{I} \text{is the pseudoscalar for the algebra, and is the highest graded blade of the algebra} [/eqn] Now we have a lot of the jargon out of the way we can continue.

First let [math] b \times c = d [/math] then our initial expression simplifies to [math] a \times d [/math] because the cross product is dual to the outer product then [math] a \times d = (a \wedge d ) ^{*} [/math] and since inner and outer products are dual [math] (a \wedge d)^{*} = a \cdot (d)^{*} = a \cdot (b \times c)^{*} = a \cdot (-b \wedge c) [/math] Next we're going to use another identity relating inner and outer products [math] a \cdot (- b \wedge c ) = -(a \cdot b)c + (-b \wedge (a \cdot c) [/math] we're almost there, now we need to take one last dual of the last term to get rid of the outer product and we finally have [eqn] -(a \cdot b) c + b (a \cdot c) [/eqn]

Now I'm going to bed.

>Inb4 mathjax fail

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>>7789502

outer product =/= exterior product

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>>7789263

Let [math] B=f \hat{i} + g \hat{j} + h \hat{k} [/math] be a continuous vector field in [math] \mathbb{R}^3 [/math].

Let [math] \Phi : \mathbb{R}^3 \rightarrow \Lambda ^1 (\mathbb{R} ^3) [/math] be the map from [math] \mathbb{R} ^3 [/math] to the space of one-forms on [math] \mathbb{R} ^3 [/math].

Let * denote the Hodge duel on [math] \mathbb{R} ^3 [/math].

Let [math] d: \Lambda ^{n} (\mathbb{R} ^3) \rightarrow \Lambda ^{n+1} (\mathbb{R} ^3) [/math] denote the exterior derivative on [math] \mathbb{R} ^3 [/math].

Finally, Let [math] \delta = (-1)^{3(n-1)+1} * \circ d \circ * [/math] be the codifferential, [math] \delta: \Lambda ^{n} (\mathbb{R} ^3) \rightarrow \Lambda ^{n-1} (\mathbb{R} ^3) [/math]

With these now specified, we can construct these classical operators.

Let [math] \bold{curl}: \mathbb{R} ^3 \rightarrow \mathbb{R} ^3 = \Phi ^{-1} \circ * \circ d \circ \Phi [/math]

Let [math] \bold{grad}: \Lambda ^{0} (\mathbb{R} ^3) \rightarrow \Lambda ^{1} (\mathbb{R} ^3) = \Phi ^{-1} \circ d [/math]

Let [math] \bold{div}: \mathbb{R} ^3 \rightarrow \Lambda ^{0} = * \circ d \circ * \circ \Phi [/math]

Let [math] \Delta: \Lambda ^{0} (\mathbb{R} ^3) \rightarrow \Lambda ^{0} (\mathbb{R} ^3) = d \delta + \delta d [/math] be the Hodge Laplacian.

Part 1/2

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>>7789668

part 2/2

First, let [math] \Phi(B)=\Phi(f \hat{i} + g \hat{j} + h \hat{k})=f dx + g dy + h dz =\beta [/math]

We are now decently close to showing [math] \boldmath{curl} \circ \boldmath{curl} (B) = (\boldmath{grad} \circ \boldmath{div} + \Phi ^{-1} \Delta \circ \Phi) (B) [/math]

Now, lets write out the sequence of operators for [math] \boldmath{curl} \circ \boldmath{curl} (B) [/math]

[math] \boldmath{curl} \circ \boldmath{curl} (B)= \Phi ^{-1} \circ * \circ d \circ \Phi \circ \Phi ^{-1} * \circ d \circ \Phi (B) \\ =\Phi ^{-1} \circ * \circ d \circ * \circ d \circ \Phi (B) [/math]

[math] = \Phi ^{-1} \circ \delta \circ d \circ \Phi (B) [/math]

[math] = \Phi ^{-1} \circ \delta \circ d (\beta) [/math]

[math] = \Phi ^{-1} \circ [-d \circ \delta +d \circ \delta + \delta \circ d] (\beta) [/math]

[math] = \Phi ^{-1} \circ [(-d \circ \delta)(\beta) +(d \circ \delta + \delta \circ d)(\beta)] [/math]

[math] = \Phi ^{-1} \circ [(-d \circ \delta)(\beta) +\Delta(\beta)] [/math]

[math] = \Phi ^{-1} \circ (-d \circ \delta)(\beta) +\Phi ^{-1} \circ \Delta(\beta) [/math]

[math] = \Phi ^{-1} \circ (d \circ * \circ d \circ *)(\beta) +\Phi ^{-1} \circ \Delta(\beta) [/math]

[math] = (\Phi ^{-1} \circ d \circ * \circ d \circ * \circ \Phi)(B) +\Phi ^{-1} \circ \Delta(\beta) [/math]

[math] = ([\Phi ^{-1} \circ d] \circ [* \circ d \circ * \circ \Phi])(B) +\Phi ^{-1} \circ \Delta(\beta) [/math]

[math] = (\boldmath{grad} \circ \boldmath{div})(B) +\Phi ^{-1} \circ \Delta \circ \Phi (B) [/math]

Where [math] -\Phi ^{-1} \circ \Delta \circ \Phi (B) [/math] is the classic vector Laplacian.

fucking rip, bold command not working :/

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>>7789671

FUCK

lol you get the picture

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>>7789668

Oops, also grad maps like $ \Lambda ^{0} \rightarrow \mathbb{R} ^3 $ not how I listed here

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>>7789679

[math] \Lambda ^{0} \rightarrow \mathbb{R} ^3 [/math]

FUCK

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>>7789711

Exterior calculus

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>>7789721

The proof of OP's problem seemed easy once he defined all of that stuff but i didn't really understand what was going on.

Could someone give me some conceptual/geometric insight to what an exterior derivative or a Hodge duel is

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>>7789263

This shit always confuses me, but I've seen this proof on stack exchange.

http://math.stackexchange.com/questions/1108020/proof-for-the-curl-of-a-curl-of-a-vector-field

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>>7789459

Bless your covariant form anon. I have not had raised and lowered indices drilled into me yet. I wish they would have taught that from the start!

Thank you for a view of notation I can only assume must be standard wherever you are! It's funny how different two people can write the same problem in a way that seems as different as Roman and moon runes.

>>7789502

Differential forms are popping up soon in physics for me. I don't know if geometric algebra is super related, but I will look into it.

>>7789502

I think I need to look at this much slower tomorrow once I'm less burnt out...

>>7789809

I looked at some other peoples' proofs, they all seem to do the same thing as me until the very last step, where they switch the derivative operators with B in a way I feel is arbitrary and lawless.

I suppose my question from the beginning was, can anyone argue why the last line of manipulation should let me swap the order of the operators and B?

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>>7789523

In geometric algebra the operation of taking a j-vector and a k-vector and combining them to make a (j+k)-vector is called the outer product. Apparently it's done to keep it distinct from the exterior product in the exterior algebra, which is related but different.

>>7789868

The two are distinct but related, for example a 2-form would just be the curl of a 1-vector field in geometric algebra.

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>>7789757

Differential forms are just a rigorous way of talking about differentials.

The exterior derivative, d, takes a function f to its differential [math]df = f_x dx + f_y dy + f_z dz[/math]. This is called a 1-form, and a general 1-form looks like [math]f dx + g dy + h dz[/math].

That doesn't tell us what dx, dy, and dz actually are, but it is an intuitive picture. We make this rigorous by thinking of a differential as a linear functional on the tangent vectors at a point, which is a useful construction because it carries over to general manifolds other than R^n (like curves and surfaces). More abstractly, one can think of differentials as sections of the cotangent bundle, but that's just an abstract way of saying "linear functional on tangent vectors".

By applying this operation again, we get 2-forms from our 1-forms, which look like [math]f dx \wedge dy + g dy \wedge dz + h dz \wedge dx[/math]. Finally, we have 3-forms, that look like [math]f dx \wedge dy \wedge dz[/math].

The hodge dual (not duel) is the map that exchanges 0-forms (smooth functions) and 3-forms, and exchanges 1-forms and 2-forms. This is done by sending f -> [math]f dx \wedge dy \wedge dz[/math], and [math]f dx + g dy + h dz[/math] to [math]f dy \wedge dz + g dz \wedge dx + h dx \wedge dy[/math].

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>>7789868

>I suppose my question from the beginning was, can anyone argue why the last line of manipulation should let me swap the order of the operators and B?

You can swap them as they're not commutative, but you also don't need to, you're almost at the goal anyway! You need to remember that there is an implicit summation over j and k, and l and m, and that summation rule sticks throughout the calculation.

Soo you sum the dj an dl operator over the first delta functions, which all disappear unless j = m and l = i, and you sum the dj and dl operators over the second delta functions which all disappear unless j=l AND you sum Bm over the delta_im function which gets you Bi. Then the result is already visible.

This is why i prefer the covariant method given here >>7789459, because it lets immediately see over which indeces you sum up.

Btw, I always liked the explanation of summation and tensor calculus in G. Wald's general relativity. If you skip the abstract defintion of a vector on a manifold (it's interesting but not necessary) he precicely explains everything about lowered and contracted indeces you need to know.

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>>7791009

Spivak's calculus on manifolds is a decent rigorous intro to differential forms as well.

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