Timeline for Beginners Guide to Cartan for Beginners [closed]
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25 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 8, 2016 at 8:33 | comment | added | Robert Bryant | @Sintrastes: It has been a while since I thought about this, but, if I remember correctly, the answer is that every $\mathcal{A}$-differentiable function in a neighborhood of $a\in\mathcal{A}$ has a Taylor series expansion of the form $$ f(x) = f_0 + f_1(x-a) + f_2(x-a)^2 + \cdots$$ where the $f_i$ for $i>1$ lie in the left-annihilator of the commutator ideal. Conversely, if $f_i$ are chosen to lie in this ideal and the above series converges, this defines an $\mathcal{A}$-differentiable function in a neighborhood of $a$. | |
| Jun 8, 2016 at 3:05 | comment | added | Nathan BeDell | @RobertBryant I'm curious to know exactly what kind of higher order functions you can make $\mathcal{A}$-differentiable (in a properly constructed algebra). Specifically, I'd like to know if there are non-commutative algebras where any convergent power series is $\mathcal{A}$-differentiable, or is that only possible in the commutative case? If not, what can we say about power series $\mathcal{A}$-differentiability$ in terms of the structure of the left-annihilator of the commutator ideal? | |
| Jan 7, 2015 at 10:00 | comment | added | Robert Bryant | (cont.) Since, for most $\mathcal{A}$, the left-annihilator of the commutator ideal is trivial, it follows that, for most $\mathcal{A}$, the only functions of the kind you want are the affine linear ones. However, you can construct many algebras $\mathcal{A}$ for which this space isn't zero, and these will have higher order functions of the kind you want. | |
| Jan 7, 2015 at 9:55 | comment | added | Robert Bryant | (cont.) To see this, you just need to compute the first prolongation of the tableau $$\Lambda = \{ L_a:\mathcal{A}\to\mathcal{A}\ \mid \ a\in \mathcal{A}\ \}.$$ Now, it is easy to show that an element $B$ of the first prolongation $$\Lambda^{(1)}=\Lambda\otimes\mathcal{A}^\ast\cap \mathcal{A}\otimes S^2(\mathcal{A}^\ast)$$ must be of the form $$B(a,b) = cab$$ where $c\in\mathcal{A}$ must satisfy $c(ab-ba)=0$ for all $a,b\in\mathcal{A}$. (Just apply the definition.) Thus, if the left-annihilator of the commutator ideal in $\mathcal{A}$ is trivial, we have $\Lambda^{(1)} = 0$. (cont.) | |
| Jan 7, 2015 at 9:50 | review | Reopen votes | |||
| Jan 7, 2015 at 22:26 | |||||
| Jan 7, 2015 at 9:49 | comment | added | Robert Bryant | (cont.) One obvious case in which there are lots of such functions is when $\mathcal{A}$ is commutative, for, in this case, any convergent power series $f(x)=f_0 + f_1 (x-a) + \tfrac12\, f_2\, (x-a)^2 + \cdots$ will satisfy the above condition (here, $f_i\in\mathcal{A}$, or, more precisely, the left multiplication operators $L_{f_i}$, must go to zero fast enough to make the series converge on some neighborhood of $a\in\mathcal{A}$). However, for most non-commutative algebras $\mathcal{A}$, this won't work, and, in fact, the affine-linear functions listed above are the only solutions. (cont.) | |
| Jan 7, 2015 at 9:39 | comment | added | Robert Bryant | @JamesS.Cook: I have been out of touch since Dec 19, so I didn't see the discussion until now. In fact, your question has a reasonable answer, but it's liable to be disappointing. Given a finite dimensional, associative $\mathbb{R}$-algebra $\mathcal{A}$, you are asking for the smooth functions $f:\mathcal{A}\to\mathcal{A}$ that satisfy $$Df(a)(bc) = (Df(a)(b))c$$ for all $a,b,c\in\mathcal{A}$. The obvious examples are $f(a) = pa+q$ for $p,q\in\mathcal{A}$ (since, in this case, $Df(a)(b) = pb$), and the question is whether there are any more and how many. You want to answer this by EDS. | |
| Dec 25, 2014 at 22:57 | vote | accept | James S. Cook | ||
| Dec 20, 2014 at 20:08 | comment | added | James S. Cook | so if there is a different way in terms of things in your thesis I am interested. | |
| Dec 20, 2014 at 20:08 | comment | added | James S. Cook | @DeaneYang Thanks for the comments, they are helpful. What I am trying to ascertain, is, if I am given a system of PDEs, let's say $n$ real PDEs in $n$ real variables, then does that system permit a reformulation as a single ODE over the algebra. Because the algebra solutions imply the general CR-equations ($n^2-n$ PDEs) I thought there would be a way to detect if the CR-equations were inconsistent with the given PDEs hence suggesting there is no natural algebra ODE which represents the given system of PDEs. I'm sorry if this is still vague, I am not in principle tied to using Tableaus... | |
| Dec 20, 2014 at 5:29 | comment | added | Deane Yang | There are other approaches to this theory. A modern cohomological approach was developed by Spencer, Quillen, Guillemin, Sternberg, and Goldschmidt. A nice description of an involutive first order differential operator was given by Guillemin and is now known as Guillemin normal form. Using this description, it is easy to construct examples of involutive systems of PDE's. Still another approach, using only basic linear algebra, is described in the first chapter of my thesis, "Involutive hyperbolic differential systems", Memoirs of the AMS. | |
| Dec 20, 2014 at 5:24 | comment | added | Deane Yang | Your question is still too vague. A tableau is essentially a way to represent a constant coefficient first order system of PDE's. So any arbitrary system of first order PDE's gives you an example of a tableau. You can then test whether the tableau is in involution or not and prolong if necessary. You need to say more about what you're hoping to learn about in the examples. | |
| Dec 20, 2014 at 3:49 | review | Reopen votes | |||
| Dec 20, 2014 at 11:09 | |||||
| Dec 20, 2014 at 3:29 | history | edited | James S. Cook | CC BY-SA 3.0 |
clarifying the question
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| Dec 20, 2014 at 3:17 | comment | added | James S. Cook | @DeaneYang The question is succinctly: does anyone have additional examples which add further insight into the Tableau method given in Chapters 4 and 5 of Cartan for Beginners. | |
| Dec 20, 2014 at 0:36 | comment | added | Deane Yang | What's the question here? | |
| Dec 20, 2014 at 0:35 | history | closed |
Ricardo Andrade Stefan Waldmann Stefan Kohl♦ Chris Godsil Deane Yang |
Needs details or clarity | |
| Dec 19, 2014 at 15:43 | answer | added | Peter Michor | timeline score: 3 | |
| Dec 19, 2014 at 14:31 | comment | added | James S. Cook | @RobertBryant thanks for the note, I have added some detail. I suppose $\mathcal{A}$-calculus has been known by various names, I'm not sure what the best name is. | |
| Dec 19, 2014 at 14:28 | history | edited | James S. Cook | CC BY-SA 3.0 |
added 774 characters in body
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| Dec 19, 2014 at 10:00 | comment | added | Robert Bryant | You need to define what you mean by $\mathcal{A}$-calculus. Without knowing this, we can't help you. | |
| Dec 19, 2014 at 2:46 | review | Close votes | |||
| Dec 20, 2014 at 0:38 | |||||
| Dec 19, 2014 at 2:36 | history | edited | James S. Cook | CC BY-SA 3.0 |
grammar
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| Dec 19, 2014 at 2:07 | history | asked | James S. Cook | CC BY-SA 3.0 |