User talk:Toby Bartels/2006

Motl vanity
I seem to remember reading somewhere that you wanted to defend User:Lumidek when his article got nominated for deletion as vanity. Well, that day has arrived. -lethe talk 11:39, 3 January 2006 (UTC)

AMA
Hello, you are receiving this message because your name is on the list of members of the Association of Members' Advocates. There is a poll being held at Wikipedia talk:Association of Members' Advocates for approval of a proposal for the revitalisation of the association. You are eligible to vote and your vote and input are welcome. Izehar 22:32, 6 January 2006 (UTC)

dear fellow christian
Hi Tony, someone removed your comment in List of Wikipedians by religion next to your name in the "catholicism" list. I reverted that. This is just to warn you it happened. --FvdP 02:39, 7 January 2006 (UTC)

V de G
Hello and Happy New Year. Some other Wikipedians and I have been making major changes/additions to Our Lady of Guadalupe, an article to which you were a prime contributor. We'd appreciate it if you'd take a look and offer criticism and/or suggestions, or even just comment on the progress. Thx, --Rockero 10:15, 8 January 2006 (UTC)

AMA Coordinator Election
Dear AMA Member,

You are entitled to vote in the AMA Coordinator election, set to begin at midnight on 3 February 2006. Please see the pages on the election and its candidates and the procedure and policy and cast a vote by e-mail!

Wally 11:37, 2 February 2006 (UTC)

Riemannian volume form
Hello Toby. I was reading an old usenet post of yours where you describe the volume pseudoform of a Riemannian manifold as the nth exterior power of the metric in the exterior algebra over the second symmetric power. I kinda like this, because I've never seen an intrinsic definition of the volume form. All the textbooks define it in terms of an orthonormal coframe, which probably can only be assumed to exist locally. However, I can't really understand the description. For starters, it seems to rely on an isomorphism between the exterior power of a symmetric power and the symmetric power of an exterior power, unless I'm misunderstanding. What's going on with that? I don't think there is such an isomorphism.

Anyway, thanks for listening. -lethe talk [ +] 18:35, 9 April 2006 (UTC)

First of all, I should tell you that I didn't fully understand the issues when I wrote that post, and I don't understand much more now. (However, I was sure that everything that I actually wrote down is correct, and —having reread the post— I'm still sure.) Accordingly, the post contains nothing like a full explanation. You're correct that orthonormal coframes can only be assumed to exist locally. And you're certainly also correct that there is no isomorphism between Altp Symk V (the pth exterior power of the kth symmetric power of V) and Symk Altp V (the kth symmetric power of the pth exterior power of V), not even in this special case (where p := n := dim V and k := 2); the dimensions don't match. (There are some interesting maps each way, I believe, but they're not relevant.)

Actually, Altn Sym2 V is never involved in this operation! Indeed, if you interpret the phrase "wedge the metric with itself n times" as a map from Sym2 V to Altn Sym2 V, then (assuming n > 1) the result is always zero, which is useless. Rather, the map should go directly from Sym2 V to Sym2 Altn V. I certainly didn't explain how that map works, and I'm not sure if I really understood it when I wrote that post. But I can explain it now: Calculating in local coordinates, if
 * An element of Sym2 V is a sum of terms of the form v ⋅ w, where ⋅ is a symmetrised tensor product.
 * More generally, an element of Sym2 Altp V is a sum of terms of the form (v1 ∧ v2 ∧ ··· ∧ vp) ⋅ (w1 ∧ w2 ∧ ··· ∧ wp).
 * To wedge an element of Sym2 Altp V and an element of Sym2 Altq V, think of the wedge product as acting factorwise on symmetric products. (This isn't really accurate, since they're different wedge products, but that's how you can think of the definition.) The result will be an element of Sym2 Altp+q V.
 * Thus, [(v1 ∧ v2 ∧ ··· ∧ vp) ⋅ (w1 ∧ w2 ∧ ··· ∧ wp)] ∧′ [(v′1 ∧ v′2 ∧ ··· ∧ v′q) ⋅ (w′1 ∧ w′2 ∧ ··· ∧ w′q)] := (v1 ∧ v2 ∧ ··· ∧ vp ∧ v′1 ∧ v′2 ∧ ··· ∧ v′q) ⋅ (w1 ∧ w2 ∧ ··· ∧ wp ∧ w′1 ∧ w′2 ∧ ··· ∧ w′q). Depending on your favourite conventions for wedge products, you might want to adjust this by p! q!⁄(p + q)!.
 * This operation is associative, but it is not alternating/antisymmetric.
 * You need to rewrite things in a manifestly symmetric fashion (balance v ⋅ w with w ⋅ v) or else use a more complicated formula.
 * I don't think that there's really anything special about k := 2 here.
 * Example: If g = dx2 + dy2, then g ∧′ g = {dx2 + dy2} ∧′ {dx2 + dy2} = {dx2 ∧′ dx2} + {dx2 ∧′ dy2} + {dy2 ∧′ dx2} + {dy2 ∧′ dy2} = {[dx ⋅ dx] ∧′ [dx ⋅ dx]} + {[dx ⋅ dx] ∧′ [dy ⋅ dy]} + {[dy ⋅ dy] ∧′ [dx ⋅ dx]} + {[dy ⋅ dy] ∧′ [dy ⋅ dy]} = {(dx ∧ dx) ⋅ (dx ∧ dx)} + {(dx ∧ dy) ⋅ (dx ∧ dy)} + {(dy ∧ dx) ⋅ (dy ∧ dx)} + {(dy ∧ dy) ⋅ (dy ∧ dy)} = {0 ⋅ 0} + {(dx ∧ dy) ⋅ (dx ∧ dy)} + {(dx ∧ dy) ⋅ (dx ∧ dy)} + {0 ⋅ 0} = 2 (dx ∧ dy)2. The 2 would disappear if one adjusted factorials; in any case, the volume pseduoform is the square root of the adjusted version (dx ∧ dy)2. This is the absolute value |dx ∧ dy|, normally written simply "dx dy".
 * $$ g = \sum_{i,j} g_{i,j} [\mathrm{d}x^{i} \cdot \mathrm{d}x^{j}] \! $$

where gi,j = gj,i, then
 * $$ g \wedge' g \wedge' \cdots \wedge' g = \mathop{n!} \det g (\mathrm{d}x^{1} \wedge \mathrm{d}x^{1} \wedge \cdots \wedge \mathrm{d}x^{n}) \! $$

where det g is the determinant of the matrix given by the gi,j. Ultimately, this result is the only reason that I know that the ideas here are correct (rather than some random useless definitions). I have no reference for this.

—Toby Bartels 22:51, 4 July 2006 (UTC)

See also this post from (much later in) the same thread. Towards the bottom of this post is a "(messy expression)" that was never writen out; I just calculated it following the principles above (adjusted by n! = 2) by hand, and I got the proper surface area element:
 * $$ \sqrt{g \wedge' g} = \sqrt{(\mathrm{d}x \wedge \mathrm{d}y)^{2} + (\mathrm{d}x \wedge \mathrm{d}z)^{2} + (\mathrm{d}y \wedge \mathrm{d}z)^{2}} \! $$
 * $$ = \sqrt{\left(\frac{\partial{x}}{\partial{u}} \frac{\partial{y}}{\partial{v}} - \frac{\partial{x}}{\partial{v}} \frac{\partial{y}}{\partial{u}}\right)^{2} + \left(\frac{\partial{x}}{\partial{u}} \frac{\partial{z}}{\partial{v}} - \frac{\partial{x}}{\partial{v}} \frac{\partial{z}}{\partial{u}}\right)^{2} + \left(\frac{\partial{y}}{\partial{u}} \frac{\partial{z}}{\partial{v}} - \frac{\partial{y}}{\partial{v}} \frac{\partial{z}}{\partial{u}}\right)^{2}}\, |\mathrm{d}u\, \mathrm{d}v| \! $$

as can be seen in (for example) Marsden's Vector Calculus (5th edition, top of page 462). —Toby Bartels

OK, so there's no exterior algebra over Sym2 V within which we are going to wedge the tensor with itself. I am very glad to learn this fact from your response, because that was the main bit on which I was stuck. Now let me see if I've understood the rest.

The wedge product is a map
 * $$ \wedge\colon \operatorname{Alt}^{p} V \otimes \operatorname{Alt}^{q} V \to \operatorname{Alt}^{p+q} V \! $$

at the next step, we are not going to obtain a map
 * $$ \wedge'\colon \operatorname{Sym}^{2} \operatorname{Alt}^{p} V \otimes \operatorname{Sym}^{2} \operatorname{Alt}^{q} V \to \operatorname{Sym}^{2} \operatorname{Alt}^{p+q} V \! $$

by arguing that Sym2 is a functor which respects the wedge product (it isn't), but rather we're going to define by fiat a new product by simply asserting that (a ⋅ b) ∧′ (c ⋅ d) = (a ∧ c) ⋅ (b ∧ d). The operation on the left-hand side is not alternating, so this isn't an element of Alt2 Sym2 V. On the other hand, the right-hand side is an element of Sym2 Alt2 V. This step is troubling me still. I mean, I follow the calculation, but what is that calculation actually doing? I'll be we have to cleverly employ some of those maps between alternating and symmetric powers that you alluded to before. Let's put that aside for the moment, and summarize: we have a map
 * $$ \operatorname{Sym}^{2} V \to \operatorname{Sym}^{2} \operatorname{Alt}^{n} V, \! $$

which you've described and given an example of in calculation. I suppose at this point I would want to check that this map is natural. Next I want an argument without local coordinates that this is a positive form and that its square root is a density on V (form which satisfies ρ(Av) = |det A|ρ(v), where v is an n-vector). Then smoothness of Alt and Sym and √ (away from zero) guarantee that this will result in a smooth section of the density bundle, and I think that finishes up the argument? -lethe talk [ +] 05:55, 7 July 2006 (UTC)

I have no reference for this. I'll bet you don't! I must have looked in a dozen different Riemannian geometry textbooks, in search of this concept. The easy books always define the form in terms of holonomic coordinates. The more sophisticated texts use the orthonormal coframe, which as far as using coordinates (i.e. a basis) is no better. And yet we know that the resulting form does not depend on the choice of coordinates. Isn't it the case that something that doesn't depend on local coordinates should admit a definition without reference to local coordinates? And aren't such definitions preferred by the tidy-minded mathematician? -lethe talk [ +] 06:01, 7 July 2006 (UTC)

I've realized a snag, trying to formalize this idea. The decomposition of the symmetric form into constituent one forms, in terms of which we defined the wedge product, is non-canonical. I think a choice of basis is required to arrive at such a decomposition. If this is the case, then this construction, while pretty, is not what I'm looking for (I'm looking for a canonical (i.e. coordinate-free) definition of the volume form). Oh well. -lethe talk [ +] 04:17, 23 July 2006 (UTC)

You certainly don't need a basis to decompose a symmetric form into constituent 1forms; I think that all you need is the ability to form partitions of unity (so the manifold must be paracompact). Of course, you must check that the result doesn't depend on how you decompose it, so in this respect it's no better than using a basis. But this is true for many other basis-independent constructions, such as the characterisation of the exterior derivative by its behaviour on sums, wedge products, and 0forms. —Toby Bartels 12:51, 24 July 2006 (UTC)

I'm a little confused by your response here. Partitions of unity are used to extend local constructions to global constructions. I'm complaining about decomposing a symmetric bilinear form into a sum of products of linear functionals (1-forms). For example, using the Gram-Schmidt process, any positive definite symmetric bilinear form may be decomposed as
 * $$ g = \sum_{i} \sigma_{i} \otimes \sigma_{i} \! $$

The existence of such a decomposition is purely a matter of linear algebra and therefore doesn't require a partition of unity. The definition of a wedge product of symmetric forms which you've given depends on the decomposition into 1-forms, and this is not canonical. I agree that partitions of unity are also noncanonical, though that's something I haven't thought about before. Is there a canonical way to define an integral? Probably not. Maybe I shouldn't worry so much about having a coordinate free description for everything. But anyway, the volume form is a local object, so it oughtn't care about something like paracompactness, right? -lethe talk [ +] 22:44, 24 July 2006 (UTC)

Sorry for the confusion; partitions of unity are indeed irrelevant. For some reason, I was thinking about forming a smooth decomposition of g on an entire manifold into a sum of symmetric squares of 1forms, but that doesn't really matter. You're right that we should only think about the linear algebra.

What do you mean by "canonical"? Although my definition of the wedge product refers to a decomposition of g into 1forms, as long as the result is independent of this decomposition, then the wedge product is canonical in any sense of the word that I know. A basis gives a convenient decomposition (which I use to verify that my definition gives the usual volume form in the end), but we can use any decomposition.

Perhaps you want a description that never makes any reference to an arbitrary choice, even if that choice makes no difference in the end? I'm not sure that this is a meaningful criterion; can't we automatically (but rather artificially) meet it by referring to the (unique) equivalence class of all decompositions of g? Not that this should stop you from looking for more pleasant descriptions, however! Even if it is a matter of taste.

—Toby Bartels 04:24, 28 July 2006 (UTC)

It's true that I've never been sure of the precise meaning of the word "canonical". I'm pretty sure I should use it to describe that dotted arrow whose existence is asserted by any universal property. Maybe also for the arrow who is a component of a natural transformation. I don't see how the way I'm using the word here fits under either of those rubrics. Nevertheless, I often see the word used to describe constructions like the one I'm seeking. A typical example is the canonical 1-form on a cotangent bundle. This 1-form is given in local coordinates by
 * $$ \theta = p_{a}\, \mathrm{d}q^{a} \! $$

(here the q are coordinates of the base space and the p are cotangent vectors, and the Einstein summation is in effect), but also has a coordinate-independent definition as
 * $$ \theta_{\sigma_{x}} = \pi^{*}\sigma_{x} \mbox{,} \! $$

where x is a point of the basespace, σ is a 1-form, and π is the projection.

If you can give me a technical sense in which the canonical 1-form is "canonical", I'd be delighted to hear it. I would expect that sense of the word to also describe the putative construction of the Riemannian volume form for which I am seeking.

These two approaches are exactly analogous to what I want to happen for the Riemannian volume pseudoform. There are plenty of other examples from differential topology/geometry: the Lie derivative, the Lie bracket, the Hodge dual, the volume form of a symplectic manifold, the volume form of a Lie group, the Levi-Civita connection on a Riemannian manifold, etc. In each case, there is a definition in terms of local coordinates, which is probably more useful for calculations, but one must check that the construction transforms covariantly under the appropriate group to justify its existence as a geometric object. The other approach is to define the construction in terms of those global data already given on the manifold (Riemannian metric, orientation, symplectic form, etc). I have somewhere picked up a heavy prejudice against the former approach and in favor of the latter.

But surely it is a matter of taste. As you point out, from a purely information-theoretic approach, one can take equivalence classes of forms, and the resulting classes will be coordinate-independent. In fact, something of this sort must be done with one of the examples I point out above; the left-invariant volume form on a Lie group is defined only up to scale, so we really only have an equivalence class of volume forms. In other words, there really is an arbitrary choice involved in that definition.

I also note that the Hodge dual is not really an example, since it relies on the Riemannian volume form for its usual geometric definition, it actually doesn't have a purely coordinate-free definition. -lethe talk [ +] 19:08, 30 July 2006 (UTC)

The strongest definition of "canonical" (as a general purpose adjective) that I know of is, as you said above, being a component of a natural transformation. I've also seen the word used in a slightly weaker sense to refer to natural transformations between functors between underlying groupoids (that is natural with respect to isomorphisms only). Since a basis is simply an isomorphism with Rn, any definition given in terms of a basis (and of other canonical conecpts) will define a canonical (in this sense) map precisely when the result is independent of the basis chosen. (But if you give a definition entirely in terms of canonical concepts, with no basis involved, then the map so defined is automatically canonical; you don't have to check basis independence.) To extend this to things that aren't even maps as such, we can fall back from "natural with respect to isomorphisms" to "functorial with respect to isomorphisms". (Naturality is actually a special case of functoriality, involving arrow categories.)

For the canonical 1-form on a cotangent bundle, the relevant categories are the category C of manifolds and the category D of manifolds equipped with a 1-form on their cotangent bundles. The construction of θ obvious assigns an object of D to each object of C; the reason that θ is canonical is that this can be extended to isomorphisms in C. That is, given a diffeomorphism φ: X → Y, (φ*)*θY = θX; in fact, this is true for φ any smooth map, so the canonical 1-form is actually functorial (at least if you define the morphisms in D correctly). We can also, with a little cleverness, view the canonical 1-form as a map, say as a map from the tangent bundle to the real line; then we don't need the (somewhat ad hoc) category D. So there are two functors from C to itself, one forming the tangent bundle of the cotenagent bundle and the other the constant functor valued at the object R (the real line); θ is canonical because, given any diffeomorphism φ: X → Y, this diagram commutes:
 * $$ \begin{matrix} \mathrm{T}\mathrm{T}^{*}X & \mathop{\longrightarrow}^{\theta_{X}} & \mathbf{R} \\ (\phi^{*})_{*}\Big\downarrow & & \Big\Vert \\ \mathrm{T}\mathrm{T}^{*}Y & \mathop{\longrightarrow}_{\theta_{Y}} & \mathbf{R} \end{matrix} \! $$

(God, I hate texvc!). Again, this diagram commutes for φ any smooth map, so this is a natural transformation, which as I remarked above is the strongest sense that I know for something to be "canonical".

As you know, I don't know a construction of the volume pseudoform on a Riemannian manifold that's defined directly in terms of well known constructions already known to be canonical. In particular, the breakdown of the metric into a sum of symmetrised products of 1-forms, while not requiring explicit reference to a basis, is not canonical; thus, one must prove that the result is independent of this breakdown to prove that the construction is canonical.

I think that you may be looking for a sense of "canonical" that means (as I just put it) "defined directly in terms of well known constructions already known to be canonical". I don't think that you can hope for this, however, at least not in any absolute sense. As you define on allegedly canonical concept in terms of other allegedly canonical concepts and so on all the way back, eventually you must return to the definition of manifold itself, which is explicitly basis (or chart) dependent. (Or if you just want to think about the linear algebra constructions, you get back to the definition of "n-dimensional vector space", which again refers to the existence of a basis). That said, you might be able to define a notion of "canonical relative to" certain very fundamental concepts (which themselves are not canonical in any absolute sense, but may be few and familiar). The field of synthetic differential geometry might be relevant here (or not).

In any case, I would use a term other than "canonical" for this, since "canonical" as I know it is much weaker:
 * canonical < basis-independent < what you're looking for

(where I don't believe in any absolute notion of "basis-independent" either).

As for the Hodge dual, I assume that you're happy with the definition of Hodge dual on a manifold equipped with a volume pseudoform; you're only unhappy with the definition of Hodge dual on a manifold equipped with a metric (a Riemannian manifold). I want to reiterate that you do have a "coordinate-independent" definition of the volume pseudoform on a Riemannian manifold, at least to the extent that this phrase has any meaning, since the breakdown of g into a sum of symmetrised products of 1-forms need not relate to any coordinate system; but I understand that the definition is not independent of all arbitrary choices. I agree that this is imperfect, even while I doubt that there is any absolute mathematical definition of perfection.

—Toby Bartels 23:17, 21 August 2006 (UTC)

While I have not fully understood your explanation of why the canonical 1-form is canonical, the explanation for why the Riemannian volume form is canonical became clear to me even before I read your post, but I have not until this moment had the wherewithal to come tell you about it; there are two functors from the category of smooth manifolds and smooth maps to, uh, the category whose objects are spaces of sections of vector bundles, whatever you want to call that. One functor assigns to each manifold the space of sections of the symmetric square bundle and the other assigns to each manifold the space of sections of the determinant bundle. There is a natural transformation which assigns to each section of the symmetric square bundle (i.e. metric) a section of the determinant bundle (i.e. a volume form).

Sounds about right, right? I haven't worked out the details, and maybe I have to restrict to the category of smooth manifolds with local diffeomorphisms.

Another thing occurred to me in the interim which I'll say at you if you've still any stomach for this conversation. It's about my search for a coordinate independent definition, our conversation about whether such a criterion is even meaningful notwithstanding.

So we've got a symmetric map from V ⊗ V → R, and we want to construct a symmetric map from AltkV ⊗ AltkV → R. This map is what we're denoting with gk (I think maybe some of my above expositions of how this map were wrong). Well actually, I've known an alternate description of this map for a long time: when we define the Hodge dual, we introduce an inner product on AltkV by
 * $$ \lang{}v_{1} \wedge \dotsb \wedge v_{k}, w_{1} \wedge \dotsb \wedge w_{k}\rang = \det (\lang v_{i}, w_{j}\rang) \! $$

This definition differs ever so slightly from yours though, at least in the way we carry out the calculation. You split the symmetric bilinear map into linear forms and then wedge them together, whereas this one splits up the k-vector argument into vectors and inner products them, taking the determinant of the resulting array of real numbers. Despite this slight difference, I'm convinced that your algorithm and this one coincide. At first I thought this definition was the resolution to all my concerns, but that lasted just a moment. For one thing, this depends on a decomposition of a k-vector into wedges of 1-vectors, the same complaint I made about your algorithm. For another thing, I object to the notion of a determinant of anything other than a linear map, which this array of numbers does not represent. So while I'm glad I recognized a familiar face in these ramblings, it still hasn't satisfied me.

I've got one hope left. It seems like the Pfaffian might be able to provide me for what I'm looking for. Well, the Wikipedia article doesn't provide a coordinate independent definition of the Pfaffian, but I think there is one. I might even be able to produce it under duress. Its property of being the square of the determinant makes it seem like it has something to say on this subject. But that route isn't clear to me; the Pfaffian applies to orthogonal transformations, not to bilinear forms, so its relation is probably only peripheral.

So that's where I am in my thinking.

Let me also respond to your comments.

You are correct about my dissatisfaction with the Hodge dual; I consider it deficient only insofar as it depends on the construction of a Riemannian volume form. I might be OK with a Hodge dual on a manifold with only a volume form, but don't I also need that inner product to finish the definition? That is
 * $$ \sigma \wedge *\tau = \lang\sigma,\tau\rang \operatorname{vol} \! $$

which needs both the Riemannian form (and its "noncanonical" in some undefined sense extension to k-forms) as well as the volume form.

You're also right, the definition of the manifold is explicitly reliant on coordinates, so the choice not to use coordinates in later definitions may be purely aesthetic or even entirely meaningless, and something which is coordinate independent doesn't care how it's defined. I found this book Natural operations in differential geometry by Kolar, Slovak, and Michor. Do you know it? It places great emphasis on naturality in differential geometry (hence the title), in particular, in the preface the authors stress that they will always define things in a coordinate independent way when possible. So I'm not alone in my prejudice for these definitions, vaguely defined though they may be. They also use category theoretic machinery quite liberally. I have not yet discovered whether they are going to give a precise (hopefully category theoretic) definition of what it means for a construction to be coordinate independent, but I'm sure that if such a defintion can be given, then they will have it. Unfortunately they do not cover Riemannian geometry at all, so I don't find a definition of Riemannian volume forms. I thought maybe to email the authors, but I can't find their email addresses.

Anyway, I'll keep thinking about it, slowly by dribs and drabs. I'll report back if I find anything good, if you care to hear it. —lethe talk [ +] 02:17, 7 September 2006 (UTC)

What a coincidence that you should mention KSM! I was just looking at it, and it reminded me of our conversation, so I came here to check if you had written anything new. I agree that it's a shame that KSM doesn't cover Riemannian geometry.

I share your preference for coordinate-free, basis-free definitions, even though I think of it as a matter of æsthetics rather than anything that I can make precise mathematically. Where we part company (with regard to our æsthetic judgments) is that (if I understand your preferences correctly) you dislike all decompositions of tensors into sums of simple tensors (not just decompositions using a basis), whereas I find using these far less objectionable than using a basis (even though I would avoid them when I can).

Although I agree that the determinant is most fundamentally a function of a linear operator, I don't see what's wrong with taking the determinant of a matrix of numbers when those numbers arise legitimately. After all, a matrix is a linear operator on Rk, which comes equipped with a canonical basis; the only question should be whether one should really have been dealing with those k2 numbers in the first place, or if one got them only by imposing a basis (or some other questionable procedure). In the case of the determinant in your definition of the inner product of simple k-vectors, these numbers are legitimate if the simple k-vectors themselves are, so the only difficulty remaining is whether we should allow ourselves to decompose arbitrary k-vectors into sums of simple ones. So I argue that you really have only one thing to object to.

I agree that your definition of this inner product agrees with mine, possibly up to some factorials that I haven't checked.

Regarding the Hodge dual, what can be defined from only a volume form is a transformation from p-vector fields to (n − p)-forms; given only a volume pseudoform, the Hodge dual is a transformation from p-vector fields to (n − p)-pseudoforms. The latter is taken (by me) as the fundamental notion of Hodge dual in the Usenet thread that you first referenced here, but of course that is not the Hodge dual that you were referring to. You're quite right that the transformation from p-forms to (n − p)-pseudoforms (which is the strongest thing that one can do given a metric but no orientation) cannot be done with a volume form or pseudoform alone.

—Toby Bartels 23:50, 7 September 2006 (UTC)

Karl Dönitz
Are you sure that the German Empire and the Weimar Republic had Länder and not provinces by some other name? --Philip Baird Shearer 22:55, 17 April 2006 (UTC)

TWF in RSS format
Hi Toby, sorry for my Engrish.

Could you help me with your professor's new format for his weekly column (This Week's Finds in Mathematical Physics). Of course I should ask him directly. But he is occupied, etc, and you seem to be very helpful to people here, so could you help me? I used to save twf in .html format (for reading off-line), but now when I save the .xml file and re-open it in Internet Explorer it says "you need template.xsl", or somehing like that. What am I doing wrong? Do you have a better suggestion for off-line reading? --Michael Dobrowski Mdob 18:41, 1 June 2006 (UTC) (Sorry, my signature is failing, dunno why...)

I don't really know how the RSS feed works or why it would have changed. But URL s like http://math.ucr.edu/home/baez/week230.html still produce fine HTML files [*], so you should still be able to download and save those, even if it's less conventient. Sorry that I can't give you any better help.

[*] Well, they're terribly non-compliant, but every browser should still read them easily.

By the way, you should write John and let him know what trouble you're having, even if you don't expect a reply, just to make sure that he's aware. That way, he'll probably fix it sometime in a few months without your ever knowing, just because he got around to it.

--Toby Bartels 05:01, 11 June 2006 (UTC)

PS: I corrected your "Engrish" as much as it seemed to need, if that helps you learn anything. And I made your signature look more or less how your talk page says that it should (although I don't understand how to make this happen automatically). --Toby

Uau!! Thanks a lot! Yes, the URL: how come I didn't think of that? So simple I even forgot to try. BTW, your URL solution to the problem is way better than the one I was thinking of. Thanks!

Thank you for correcting my bad English spellings too. Oh and I've fixed my signature with the help of this page. Just enable "Raw signatures". Voila: Mdob | Talk 19:34, 18 June 2006 (UTC)

I'm glad it worked out. --Toby Bartels 18:37, 3 July 2006 (UTC)

Your AMA statement was deleted
Your AMA statement was deleted apparently carelessly from the AMA statements page (see this edit)if your statement needs to be restored, the text is at User:Pedant/AMA error Pedant 02:03, 22 July 2006 (UTC)

On the board
Hard to say... I'm not a fan of Erik Moller being on the board, mainly because I disagree with a lot of his methods; on the other hand, it may be appropriate for someone who already wields disproportionate power to be officially recognized. Otherwise, it's hard. My uninformed bias would be towards the non-English users who have been around for a while; I like Oscar's candidacy (based simply on his presentation) -- he seems to be the right type of user for the board, a polylinguist polymath who sees the big big picture. --The Cunctator 04:01, 15 September 2006 (UTC)