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This answer says that in surreal numbers $\ln \omega=\omega^{1/\omega}$.

At the same time, this Wikipedia article says that transseries $\mathbb{T}^{LE}$ are isomorphic to a subfield of $No$ with its natural field structure and operations and series $x$ corresponding to the surreal number $\omega$.

Also, this article says that Hardy fields are isomorphic to $No(\omega_1)$, also a subset of surreals.

But in transseries, and in Hardy fields $\ln x\ne x^{1/x}$. How this could be the case if they are isomorphic to subfields of surreals?

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    $\begingroup$ Are you sure these articles are referring to isomorphism as [ordered] exponential fields, rather than merely isomorphism as [ordered] fields? I would expect the answer here is just "they're isomorphic as [ordered] fields, not as exponential fields". $\endgroup$
    – Harry Altman
    Commented May 7 at 4:01
  • $\begingroup$ @HarryAltman hmm, does not isomorphism of fields also mean exponentiation and logarithms should coincide? There are power wseries, etc... $\endgroup$
    – Anixx
    Commented May 7 at 9:40
  • $\begingroup$ No. For one thing, even if the exponential in all these contexts were defined by the usual power series, isomorphism merely as fields rather than as ordered or topological fields wouldn't imply much, because the isomorphism wouldn't have to preserve the topology, so the series could converge to different things on both sides. But also, the exponential isn't defined that way in the surreals in the first place! It's not defined by a power series there. It can't be, because there is no nontrivial convergence of sequences in the surreals! So even preserving the order/topology wouldn't do it here. $\endgroup$
    – Harry Altman
    Commented May 7 at 19:08
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    $\begingroup$ @HarryAltman This paper says the transseries are embeeded into No as exponential ordered field math.ucla.edu/~matthias/pdf/BM6.pdf (i.imgur.com/AFP0RQ2.png) $\endgroup$
    – Anixx
    Commented May 7 at 20:18
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    $\begingroup$ Interesting! But yeah what nombre says below is true, I didn't even account for that... $\endgroup$
    – Harry Altman
    Commented May 7 at 22:04

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The exponentiation in $\omega^{\frac{1}{\omega}} = \ln \omega$ is not that given by the exponential function. It is the value at $\frac{1}{\omega}$ of Conway's $\omega$-map.

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  • $\begingroup$ Can you explain please? It is not exponentiation? Why it is written as exponentiation then? Can it be represented in other forms? I thought Conway form is akin to a power series. $\endgroup$
    – Anixx
    Commented May 7 at 20:20
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    $\begingroup$ You could define exponentiation on the surreals by $a^b = \exp(b\log a)$, but if you do this, you'll find that $\omega^x$ does not agree with the operation normally called $\omega^x$, which is, as nombre says, Conway's $\omega$-map. Why is it written that way? Because it's convenient, and it's older and more commonly used than the other sort. Math is full of overloaded notation. Yes, it's confusing and unfortunate, but you can't focus on how things are written. $\endgroup$
    – Harry Altman
    Commented May 7 at 22:06
  • $\begingroup$ @HarryAltman so $\ln \omega$ is just germ of $\ln x$, nothing more and nothing less, if we are given that Hardy field or transseries embedding. $\endgroup$
    – Anixx
    Commented May 7 at 22:12
  • $\begingroup$ @HarryAltman $\log a$ is also defined in the sense of Conway map. $\endgroup$
    – Anixx
    Commented May 7 at 22:44
  • $\begingroup$ The things you're saying don't make sense. The Conway map, unlike the exponential, isn't surjective (onto the positive surreals). I think you need to go and just, like, read up on surreals some more. Not sure MathOverflow is really the place to be asking these questions. $\endgroup$
    – Harry Altman
    Commented May 7 at 23:04

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