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Is the multiplicative Group of surcomplex numbers of modulus 1 isomorphic to the additive Group of the surreal numbers modulo the sub-Group of surreal integers? And, do Norman Alling's surreal extensions of sine and cosine (defined in section 7.5 of his book "Foundations of analysis over surreal number fields") accomplish the isomorphism?

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Let me say at least this: the usual series for sine and cosine "converge" for the finite surreals, and provide an isomorphism from (the finite surreals modulo the standard integers) onto (the surcomplex unit circle).

An alternate for the sine on the finite surreals, write $x = a+z$ where $a$ is a standard real and $z$ is infinitesimal, then use the addition formulas for $\sin(a+z)$ and $\cos(a+z)$.

added March 18 Extension to all surreals depends on the choice for the complementary subgroup of the finite surreals. What (beyond the usual $\mathbb Z$) should be called an "integer". Conway has such a choice in his formulation, called $\mathbf{Oz}$.

surjective ... Conway emphasizes more the algebraic and combinatorial side, less the analytic side. But, in fact, this same thing will work in all the usual canonical ways of constructing nonarchimedean extensions of the reals.

In nonstandard analysis, $\sin$ and $\cos$ have corresponding nonstandard versions, and surjectivity is a first-order property, so it transfers.

In transseries, there are many possibilities: series expansion for $\arcsin$; an integral; a solution of a differential equation; ...

In the surreals, Erlich [LINK] showed $\mathbf{No}$ can be realized as a space of Hahn series, and after that it will be the same as for transseries. It does seem less convenient in Conway's original formulation, admittedly.

added March 19
Here is how we do it when using Hahn series. Once you reach a certain point in Conway's book ONAG, you can do this also for surreals, using his Theorem 23 with his "normal forms".

Hahn series look like $\sum_{i \in I} c_i g_i$, where the coefficients $c_i$ are real, and the "monomials" $g_i$ are reverse well-ordered. One possible monomial is $1$; monomials larger than that are "infinite", those smaller are "infinitesimal". The set of possible monomials is an ordered abelian group under multiplication.

Given a general element $A$ of our field of Hahn series, we write it as $A = L + t + S$, where every monomial in $L$ is infinite, $t \in \mathbb R$, and every monomial in $S$ is infinitesimal. Define $$\begin{align} \sin A &= \sin t \cos S + \cos t \sin S, \cr \cos A &= \cos t \cos S - \sin t \sin S \end{align}$$ and for infinitesimal $S$, $$\begin{align*} \sin S &= S - \frac{1}{6} S^3 + \frac{1}{5!} S^5 + \dots, \cr \cos S &= 1 - \frac{1}{2} S^2 + \frac{1}{4!}S^4 + \dots, \end{align*}$$ with convergence in the most trivial sense: each monomial occurs in only finitely many terms of the expansion, so you just collect terms. Then observe that there is an inverse series: $$ \arcsin T = T + \frac{1}{6} T^3 + \frac{3}{40} T^5 + \dots $$ with convergence in the same sense. Actually, for the surjectivity in this problem, it may be more convenient to use one series $\arctan T$ rather than two series $\arcsin$ and $\arccos$. So: Given $X,T$ with $X^2+Y^2=1$ we claim there is $A$ with $\sin A = X, \cos A = Y$. We should take either $A = \arctan Y/X$ or that plus $\pi$, depending on the signs of $X$ and $Y$.

This is getting to be too long for an answer...

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  • $\begingroup$ How do we know the map is surjective? $\endgroup$ Commented Mar 18, 2012 at 2:47
  • $\begingroup$ Thanks for the remark about the choice of "complementary subgroup"; it clarified things for me. But I still don't see the surjectivity of the map from the surreals to the surcomplex unit circle. I agree that in non-standard analysis there's a transfer principle, but surreal analysis is different from NSA (I'll start a new thread on this: mathoverflow.net/questions/91646/…), and in any case it's not clear to me that bracket-based definitions of sin and cos are first-order. $\endgroup$ Commented Mar 19, 2012 at 18:11
  • $\begingroup$ I don't even know what bracket-based definitions of sin and cos are. In ONAG Conway only remarks that the obvious ones don't work. $\endgroup$ Commented Mar 19, 2012 at 19:56
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The answer to the first question is yes and the answer to the second question is no. As Ovidiu Costin confirmed in an email to me, the desired isomorphism can be constructed using an idea I learned from him regarding how to define sin/cos on all the surreals. The idea in Ovidiu's words follows, where N ranges over the omnific integers (finite and infinite).

With sin/cos the idea is not mine but Martin's (or it even goes back to Conway). What it gives is the following prescription: sin(2 pi N+delta)=sin(delta), if delta\in [0,2\pi). This can be taken as a definition as well. Similarly with cos. Clearly sin/cos are well defined on all surreals. Any isomorphism should now be straightforward.

Regards,

Philip Ehrlich

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  • $\begingroup$ I don't understand; say N=1. Then the prescription sin(2 pi N delta) = sin(delta) doesn't hold. Maybe N can only be infinite? In any case, I'd like to see more details. $\endgroup$ Commented Mar 16, 2012 at 18:30
  • $\begingroup$ Presumably: sin(2 pi N + delta)=sin(delta) $\endgroup$ Commented Mar 16, 2012 at 21:10
  • $\begingroup$ Oops, it should have read: sin(2 pi N+delta)=sin(delta) not "sin(2 pi N delta)". Sorry! I have now corrected the typo in the original answer. Philip Ehrlich $\endgroup$ Commented Mar 16, 2012 at 21:30
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    $\begingroup$ Does one prove surjectivity using power series expansions of arcsine and arccosine? If not that way, then how? $\endgroup$ Commented Mar 18, 2012 at 2:50
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The following two questions were asked:

1: Is the multiplicative Group of surcomplex numbers of modulus 1 isomorphic to the additive Group of the surreal numbers modulo the subgroup of surreal integers?

2: Does Norman Alling’s surreal extensions of sin and cos (defined in his book) accomplish the isomorphism.

In my earlier posting I said the answer to 1 is yes and the answer to 2 is no.

In response to the request for further details, first note that, as Alling himself observes, his extensions of the definitions of sin and cos via series only applies to infinitesimals. Accordingly, we need to know that sin and cos are well defined throughout the surreals. This is the import of Ovidiu Costin’s observation (taught to him by Martin Kruskal) that one can define sin(2 pi N+delta)=sin(delta), if delta is in [0, 2pi) (and analogously for Cos) where N ranges over all the omnific integers (finite and infinite). Hence, my answer to 2.

As to the isomorphism itself, note that since the properties of sin, cos are the same for real as for surreal numbers, one would simply write that (x+iy) with x, y in [-1,1] and x^2+y^2=1 is mapped to theta where there is a unique theta such that cos(theta)=x, sin(theta) =y. For theta in [0, 2pi], both sin(theta) and cos(theta) can be defined (following Kruskal) in terms of a surreal loop bracket { | } containing upper and lower truncates of the usual Taylor series (using the ideas found on pp. 145-146 of Gonshor’s book on surreal numbers). Alternatively, one can skip the use of surreal loop brackets and proceed as follows: for all surreal x, write x = 2pi N+r+delta, where N is an omnific integer, r is a real and delta is an infinitesimal and define sin(x)=sin(r)cos(delta)+cos(r)sin(delta)

and

cos(x)=cos(r)cos(delta)-sin(r)sin(delta),

where sin(r) and cos(r) are the usual sin and cos, and sin(delta) and cos(delta) are defined in terms of Taylor series.

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    $\begingroup$ I'm afraid I still don't get it. Philip writes: "since the properties of sin, cos are the same for real as for surreal numbers, one would simply write that (x+iy) with x, y in [-1,1] and x^2+y^2=1 is mapped to theta where there is a unique theta such that cos(theta)=x, sin(theta)=y". But why are the properties of sin and cos the same for real as for surreal numbers? If there is a transfer principle at work here, I'd like to see a clear statement of it. $\endgroup$ Commented Mar 19, 2012 at 17:34

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