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Does pure mathematics bring a good, concise, straightforward, standard notation to express a symmetry argument?

I will give an example but as you can see, the example is too long—which is exactly my point. Is there not a much shorter, clearer, more conventional way to write something like this?

EXAMPLE

For example, per Parseval,

$$\sum_{j=0}^{10} e^{i2\pi(3/11)j} = 0$$

or more generally,

$$\sum_{j=0}^{N-1} e^{i2\pi(n/N)j} = 0$$

as long as $N\neq 0$; as $n$ is not a multiple of $N$; as $(n,N)\in\mathbb Z$, etc.

There exist several not-very-interesting ways one could prove Parseval right, for example by evaluating the above expression or, if you prefer, by evaluating the perturbed

$$\lim_{\epsilon\rightarrow 0^{+}}\sum_{j=0}^{N-1} e^{(i-\epsilon)2\pi(n/N)j} = 0$$

as a geometric series—but as I said, those ways aren't very interesting because they miss Parseval's point.

APPEAL TO SYMMETRY

Parseval is interesting not because he can add up a series but because he makes a statement about symmetry. Parseval says that a point in every direction is no point at all.

The sample argument above may be correct but, insofar as its method evokes no mental image of symmetry, I think it a poor argument. Does mathematics not bring some kind of symmetry symbology to appeal to symmetry more directly?

In the example, does mathematics not bring some kind of reasonably simple notation to say, "a point in every direction is no point at all"?

BACKGROUND

For information, my mathematical level is this: I hold a master's degree in engineering; have read a book on real analysis; have taken an undergraduate math-department course in complex analysis; and have taken two graduate math-department courses in linear algebra. I don't mind if you give an answer that flies over my head, but if you start talking in a terse manner about abstract algebra, rings, Lie groups, etc., you'll probably lose me.

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Many symmetry arguments have geometry underlying them, and that's the case with the proof of the equation $$\sum_{j=0}^{10} e^{i2\pi(3/11)j} = 0 $$ In the complex plane, the numbers $e^{i2\pi(3/11)j}$ for $j=0,...,10$ can be depicted by subdividing the unit circle at $11$ equally spaced points starting with $e^{i2\pi(3/11)0}=1 + 0i$, and then drawing vectors from the origin to each of those $11$ points.

Here comes the symmetry part: those $11$ vectors are symmetric with respect to rotation of the plane through angle $2\pi/11$.

Denote those $11$ vectors as $v_1,...,v_{11}$. Each of them has length $1$ (because they are radii of the unit circle).

Next, draw an equilateral $11$-gon of side length $1$ having one side parallel to $v_1$ denoted $e_1$, and centered at some point $C$ (I don't care which point).

Here comes the second symmetry part: that $11$-gon is symmetric with respect to rotation of the plane about $C$ through angle $2\pi/11$.

Draw an arrow on $e_1$ indicating its direction that is parallel to the direction of $v_1$. Now continue that direction around to each of the 11 sides of the $11$-gon.

You can now make a one-to-one correspondence between the vectors $v_1,...,v_{11}$ and the sides of $C$, listing those sides as $e_1,...,e_{11}$, so that $e_k$ is parallel to $v_k$. In other words, the vectors $e_k,v_k$ are identical.

Since the vectors $e_1,...,e_{11}$ close up to form the boundary of the 11-gon, it follows that their vector sum equals zero. Hence, the vector sum of $v_1,...,v_{11}$ equals zero.

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  • $\begingroup$ I see. So the way to make a symmetry argument is to establish invariance during a transformation—as one would, for example, establish that a function were odd or even? That makes sense. +1 $\endgroup$ – thb Jun 19 '17 at 22:18
  • $\begingroup$ Right. Even (odd) symmetry is invariance under an order 2 symmetry group consisting of a reflection of the plane across a line (point). And the symmetry underlying your question is an order $11$ rotational symmetry group in the complex plane. $\endgroup$ – Lee Mosher Jun 20 '17 at 13:53
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By De Moivre's formula, the complex numbers you are adding are the roots of $z^{N}-1=0$, and this polynomial has trace equal to zero.

Remark. The sum of roots (=trace) of a polynomial of degree $N$ (i.e., the first elementary symmetric function of the roots) is minus the coefficient of the term of degree $N-1$.

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