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It's well-known that $\sin(\pi \frac pq)$ is always algebraic. In particular, as I understand, it can always be expressed in terms of radicals, because it can be connected to the abelian group of $e^{i\pi \frac pq}$. Because these can always be expressed in terms of radicals -- this means there must be some other algebraic numbers (for instance, the roots of irreducible quintics) whose inverse sines are not a rational multiple of pi. So what form do these take? Is $\sin(\sqrt 2 \pi)$ known to be transcendental, for instance? That would be the "first place to start looking", to me, but I have little basis for that.

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  • $\begingroup$ You are wrong about $\sin{\pi \frac pq}$ always being expressible in radicals. Please see this example with $\pi/7$ mathworld.wolfram.com/TrigonometryAnglesPi7.html $\endgroup$ – Tyrell Jan 12 '17 at 15:44
  • $\begingroup$ If e^ix is transcendental, at least one of its real or imaginary parts is transcendental, because otherwise we could add a+bi and write e^ix as a sum of two algebraics. Since the absolute value is 1, a=sqrt(1-b^2), so if one part is algebraic so is the other. Thus since e^ix is transcendental, both its complex parts are. And since e^ix-e^-ix is the imaginary part, we're done $\endgroup$ – Alex Meiburg Jan 13 '17 at 7:54
  • $\begingroup$ Why the hell I didn't see that? Thanks. $\endgroup$ – Tyrell Jan 13 '17 at 8:10
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$\sin \pi \alpha$, for $\alpha$ irrational but algebraic, is transcendental by the Gelfond-Schneider theorem.

To start, write

$$\sin \pi \alpha = \frac{e^{i \pi \alpha} - e^{-i \pi \alpha}}{2i}.$$

We will actually show that $e^{i \pi \alpha}$ is transcendental. This follows from writing it as $(e^{i \pi})^{\alpha} = (-1)^{\alpha}$. (This notation may look funny; for complex numbers, $a^b$ is multivalued, and Gelfond-Schneider applies to all of the possible values.)

So I don't know what there is to say beyond "take the inverse sine of some algebraic number, then divide by $\pi$."

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    $\begingroup$ @Alex: no, there are lots of obstructions. First, expressions of that form are always real. Second, they are always between $-1$ and $1$. Third, by construction they are half of a sum of two roots of unity, which is an algebraic integer (en.wikipedia.org/wiki/Algebraic_integer, and for example $\frac{1}{3}$ is not half of an algebraic integer. Fourth, they are always half the sum of two roots of unity, and it's not hard to write down numbers which can be written as the sum of, say, three roots of unity but not two. Fifth, they are always the sums of roots of unity, and... $\endgroup$ – Qiaochu Yuan Mar 21 '15 at 8:07
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    $\begingroup$ ...so their Galois groups are abelian, and there are algebraic numbers expressible by radicals whose Galois groups are solvable but not abelian, e.g. roots of cubics with Galois group $S_3$. Sixth... etc. $\endgroup$ – Qiaochu Yuan Mar 21 '15 at 8:07
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    $\begingroup$ The closest approximation of that statement by a true statement that I can think of is the Kronecker-Weber theorem (en.wikipedia.org/wiki/Kronecker%E2%80%93Weber_theorem), which implies that if the Galois group of an algebraic number is abelian (this is stronger than saying it can be expressed by radicals) then it is a rational linear combination of roots of unity. $\endgroup$ – Qiaochu Yuan Mar 21 '15 at 8:10
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    $\begingroup$ The fourth obstruction above is a bit silly because the easiest way to make it precise reduces to the first or second obstructions. Let me offer a different one: not only is $\sin \pi \frac{p}{q}$ between $-1$ and $1$, but so are all of its conjugates (en.wikipedia.org/wiki/Conjugate_element_%28field_theory%29). On the other hand, $\sqrt{2} - 1$ is an algebraic number (even an algebraic integer) between $-1$ and $1$ expressible by radicals, but its conjugate $\sqrt{2} + 1$ is not between $-1$ and $1$. $\endgroup$ – Qiaochu Yuan Mar 21 '15 at 8:16
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    $\begingroup$ Nevertheless, there is an interesting generalization of the idea of using a transcendental function like the exponential function to produce interesting algebraic numbers, namely Kronecker's Jugendtraum (en.wikipedia.org/wiki/Hilbert%27s_twelfth_problem). $\endgroup$ – Qiaochu Yuan Mar 21 '15 at 8:22

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