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February 28

Conservative field in high dimension

Given a bounded, continuously differentiable vector field on a bounded simply connected domain in ${ℝ}^n$ (where n>3); How can I prove that this field is conservative? Differentiating the field gives the Hessian matrix of the potential (if it exists). Clearly it is a necessary condition that the Hessian is symmetric, and for n ≤ 3 a symmetric Hessian is equivalent to zero curl, so in that case it is also a sufficient condition. However, curl is not defined for n>3, and I can't find any citeable reference that will tell me if a symmetric derivative matrix implies a conservative field. Any help will be appreciated. PeR (talk) 09:07, 28 February 2014 (UTC)

This can be formulated in the language of differential forms. Instead of your vector field $F=(F_1,\dots ,F_n)$, you look at the 1-form $v_F=F_{1}d{x}_1+F_{2}d{x}_2+\dots +F_{n}d{x}_n$. By Poincaré's lemma, if F is defined on all of ${ℝ}^n$, $v_F$ is exact if and only if it is closed. Closed means $d{v}_F=0$, which is (look at exterior derivative) another way of writing ${\partial }_{x_i}{F}_j={\partial }_{x_j}{F}_i$. Exact means there is a 0-form (which is the same as a function) U such that $v_F=dU$, which is the same as $F=\mathrm{\nabla }U$.

So a "symmetric derivative matrix" implies that the field is conservative, provided the domain on which it is defined is contractible (although simply connected is enough).

I hope that was not too confusing. There are elementary treatments of the fact (and they are not difficult), but all citable references I have on my desk at the moment are in German. For example, H. Heuser, Lehrbuch der Analysis 2, 8th edition, Stuttgart: Teubner 1993, Satz 182.2. Basically, the proof is the same as for two or three dimensions. —Kusma (t·c) 09:53, 28 February 2014 (UTC)

Great! Thank you so much for the explanation. PeR (talk) 08:09, 1 March 2014 (UTC)

Think you can find most of it in Introduction to smooth manifolds by John M. Lee YohanN7 (talk) 03:52, 3 March 2014 (UTC)

Sum of Reciprocal (prime) powers.

Let D be the set of integers > 1. Let E be the set of Prime Numbers (2,3,5,7,11...).

• W = (sum over s in D, sum over t in D, (1/(s^t))
• X = (sum over s in D, sum over t in E, (1/(s^t))
• Y = (sum over s in E, sum over t in D, (1/(s^t))
• Z = (sum over s in E, sum over t in E, (1/(s^t))

Any ideas on how to prove any of these being finite or infinite. Note that W>Y>Z and W>X>Z.Naraht (talk) 19:14, 28 February 2014 (UTC)

Using the sum of a geometric series,

$W=\sum _{n\ge 2}\sum _{k\ge 2}\frac{1}{n^k}=\sum _{n\ge 2}\frac{1}{n^2(1-\frac{1}{n})},$

and that sum converges. So all of your sums converge. —Kusma (t·c) 20:06, 28 February 2014 (UTC)

But isn't $\sum _{n\ge 2}\frac{1}{n^2(1-\frac{1}{n})}$ =1?

— Preceding unsigned comment added by Naraht (talk • contribs) 23:21, 1 March 2014‎

Indeed it is, since $\frac{1}{n^2(1-\frac{1}{n})}=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$. So $W=1$. —Kusma (t·c) 06:44, 2 March 2014 (UTC)

Any information on the values for X, Y, or Z or which is greater, X or Y?Naraht (talk) 04:48, 3 March 2014 (UTC)

I don't know anything about these values. For Y, the closest thing I can find is the "integral" section in Prime zeta function. —Kusma (t·c) 19:54, 4 March 2014 (UTC)

Thank you.Naraht (talk) 20:04, 4 March 2014 (UTC)