What is the value of the critical temperature?

Question

What is the value of critical temperature in the 2D classical Ising model?

My Understanding
Suppose one can write the partition function for the 2D classical Ising model in high-temperature expansion and low-temperature expansion.
In High-temperature expansion, the partition function has form:
begin{align*}
{cal Z}_{HTE}&=2^Ncosh K^{2N}left(1+Ntanh K^4+2Ntanh K^6+dotsright)
    &=2^Ncosh K^{2N}f(tanh K)
end{align*}
While in low-temperature expansion, the partition function has a form:
begin{align*}
{cal Z}_{LTE}&=e^{2NK}left(1+Ne^{-4times 2K}+2Ne^{-6times 2K}+dotsright)
    &=e^{2NK}f(e^{-2K})
end{align*}

"Yang-Lee theorem states that: Yang-Lee zeros or the zeros of partition
function lie on an imaginary complex plane. In the thermodynamic
limit, the point where zeros cross the real axis marks the phase
transition point." In partition function above, if the Yang-Lee zeros
crosses the real axis at the critical point, then zeros must be
contained in function $f$ at point $K_c$ such that

$$e^{-2K_c}=tanh K_c$$
with simple algebra we get
$$ e^{-4K_c}+2e^{K_c}-1=0implies e^{-2K_c}=-1pmsqrt{2}$$
Only the positive solution is acceptable, which leads to relation
begin{equation}label{kwrelationeq}
 k_BT_c=frac{2J}{ln(1+sqrt{2})}backsimeq 2.269J
end{equation}

My Question
The statement I made up about Yang-Lee zeros is wholly made up by myself. However, I am not 100% sure that if I am going in the right direction. So, is the statement quoted above is correct?

When I searched about Kramers–Wannier duality relation, I always got to how to write the partition function. However, the explanation about duality was not clear to me. So I tried to link with Zeros of the partition function.

Yvan Velenik · Accepted Answer

Let $phi(beta)$ denote the free energy density at inverse temperature $beta$ (and with no magnetic field). Then, the argument you sketch (a rigorous version of which can be found in Section 3.10.1 of this book) implies that
$$
phi(beta) = phi(beta^*) - log sinh(2beta^*),
tag{$star$}
$$
where
$$
beta^*=mathrm{arctanh},(e^{-2beta}).
$$
This means that the free energy is essentially invariant under the transformation
$betamapstobeta^*$, which interchanges the low and high temperature regimes: this function is an involution with a unique fixed point $beta_{rm sd}=tfrac12log(1+sqrt{2})$, which interchange the intervals $[0,beta_{rm sd})$ and $(beta_{rm sd},infty]$.
Since $betamapstobeta^*$ and $betamapsto log sinh(2beta)$ are both analytic functions on $(0,infty)$, it
follows from ($star$) that any non-analytic behavior of $phi$ at some inverse temperature $beta$ must also imply a non-analytic behavior at $beta^*$.
Consequently, if one assumes that the function $betamapsto phi(beta)$

is non-analytic at $beta_{rm c}$,
is analytic everywhere else,

then $beta_{rm c}$ must coincide with $beta_{rm sd}$.

Note that, above, the free energy is seen as a function of $beta$ (or, if you wish, the coupling constant $K$ in your notation). The magnetic field is always equal to $0$.
The Lee-Yang theorem, however, is about the analyticity of the free energy as a function of the magnetic field $h$. Thus, it cannot help you with establishing the two assumptions used in the argument above.

The first computation of $beta_{rm c}$ was done by Onsager in his famous work. There exist, however, several alternative ways to prove that the critical and self-dual points coincide without using the exact solution. These approaches have the advantage of being far more robust. They can, for instance, be applied to more general planar graphs or to other models such as the Potts model. You can find more about that in these lecture notes. Note that these approaches do not proceed along the lines above (that is, they do not rely on the 2 assumptions made above). Of course, duality arguments still play the central role.

What is the value of the critical temperature?

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