Real forms of $\mathbb{G}_m$ as a variety

Question

Notation: Denote $k=\mathbb{R}$, $K=\mathbb{C}$ and let $G:=\text{Gal}(K/k)=\{\text{id},\sigma\}$, where $\sigma$ is complex conjugation.

Background: In the Notices article A Gentle Introduction to Arithmetic Toric Varieties the author introduces (in slightly different notation) three varieties over $k$:

• $\mathbb{G}_m=\{uv-1=0\}$,
• $\text{S}^1=\{x^2+y^2-1=0\}$,
• $Y=\{a^2+b^2+1=0\}$.

He establishes that they are pairwise non-isomorphic: e.g. looking at the real points in Euclidean topology $\mathbb{G}_m(k)$ has two connected components, while $\text{S}^1(k)$ has only a single connected component and $Y(k)=\emptyset$ has no $k$-points at all.

He then explains that $\mathbb{G}_m$ and $\text{S}^1$ are twisted forms of each other, since they become isomorphic (as affine $K$-varieties) after base change to $K$, explicitly: $$\varphi:\text{S}^1\times_kK\rightarrow\mathbb{G}_m\times_kK,\quad (x,y)\mapsto(u,v)=(x+iy,x-iy).$$ Explicitly the inverse morphism is given by: $$\varphi^{-1}:\mathbb{G}_m\times_kK\rightarrow\text{S}^1\times_kK,\quad (u,v)\mapsto(x,y)=\left(\frac{u+v}{2},\frac{u-v}{2i}\right).$$ As far as I see $Y$ also becomes isomorphic to $\mathbb{G}_m$ and $\text{S}^1$ after base change to $K$, an isomorphism is given e.g. by: $$\psi:\text{S}^1\times_kK\rightarrow Y\times_kK,\quad (x,y)\mapsto(a,b)=(ix,iy),$$ $$\psi^{-1}:Y\times_kK\rightarrow\text{S}^1\times_kK,\quad (a,b)\mapsto(x,y)=(-ia,-ib).$$ This would indicate that the complex affine variety $\mathbb{G}_m\times_kK$ has (at least) 3 distinct real forms, namely $\mathbb{G}_m, \text{S}^1$ and $Y$. Note that I do not consider the structure of a group variety, with which you could endow $\mathbb{G}_m, \text{S}^1$ in the usual way but not $Y$ as it is pointless (over $k$).

As is well known for classes of varieties $X$ that form a ‘stack’ (e.g. quasi-projective varieties - so affine $X$ is OK) $k$-forms of some $X\times_kK$ are in bijection to $$\text{Tw}(X)=\text{H}^1_{\text{gal}}(k,\text{Aut}_{\text{var}}(X\times_kK))=\text{H}^1_{\text{grp}}(\text{Gal}(K/k),\text{Aut}_{\text{var}}(X\times_kK))$$ (cf., e.g., Gille/Szamuely, "Central Simple Algebras and Galois Cohomology", Thm. 2.3.3 and sec. 5.2 for special versions). The author claims $$\text{Tw}(\mathbb{G}_m)=\{\mathbb{G}_m, \text{S}^1\}$$ only contains 2 distinct elements.

And indeed using this post (and using that morphisms of affine varieties are in bijection to algebra morphisms) we see that $\text{End}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}\right\}$ and therefore $A:=\text{Aut}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\cong K^{\times}\times\mathbb{Z}/2\mathbb{Z}$. The Galois action of $\text{G}$ on $A$ is induced by $\tilde{\sigma}:A\rightarrow A, (a,\pm 1)\mapsto(\sigma(a),\pm 1)=(\bar{a},\pm 1)$ using the identification with $K^{\times}\times\mathbb{Z}/2\mathbb{Z}$ just given.

Using the standard method to compute group cohomology for cyclic groups through norm and difference maps, $N=\sum\limits_{i=0}^{2-1}\tilde{\sigma}^i=\tilde{\sigma}+\tilde{\text{id}}$ and $\Delta=\tilde{\sigma}-\tilde{\text{id}}$ (cf., e.g., Gille/Szamuely, Ex. 3.2.9), as $$\text{H}^1_{\text{grp}}(\text{G},A)\cong\text{ker}(N)/\text{im}(\Delta)$$ this can easily seen to indeed be $\cong\mathbb{Z}/2\mathbb{Z}$.

Question: Since $3>2$ there must be a mistake in these considerations: What is this mistake and what is $\text{Tw}(\mathbb{G}_m)$?

Remark 1: As indicated above one could also look at the additional structure of group varieties. For this variant we have, if I am not mistaken, $B:=\text{Aut}_{\text{grp-var}}(\mathbb{G}_m\times_kK)\cong\left\{1\cdot t^n:n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\cong \mathbb{Z}/2\mathbb{Z}$ which yields $\text{Tw}_{\text{grp-var}}(\mathbb{G}_m)=\text{H}^1_{\text{grp}}(\text{G},B)\cong\mathbb{Z}/2\mathbb{Z}$. Since $Y$ has no $k$-point it cannot have the structure of a group variety, thus $Y$ is no real form of the complex group variety $\mathbb{G}_m\times_kK$ and there is no contradiction from the above arguments. The issue therefore lies in classifying real forms of $\mathbb{G}_m\times_kK$ as only affine $k$-varieties without any (potential) group structure.

Remark 2: I choose the notation $K/k$ to indicate that analogous issues should also arise for e.g. $\mathbb{Q}(i)/\mathbb{Q}$ the Gaussian rationals and probably even in positive characteristic (probably $\neq 2$, since $\varphi^{-1}$ requires inversion of $2$).

Answer 1

I will post this answer just as a complement to Thomas Preu's nice computation above.

Let us write $\Gamma:=\mathrm{Gal}(\mathbf{C}/\mathbf{R})$. Also, let me model $\mathbf{Z}/2\mathbf{Z}$ as $\{\pm 1\}$ -- I will identify them in what follows. As already observed we have a short exact sequence of $\Gamma$-groups

$$1\to \mathbf{C}^\times\to \mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}})\to \mathbf{Z}/2\mathbf{Z}\to 1,$$

where the latter term has trivial $\Gamma$-action and the first the usual $\Gamma$-action. We then know that we get (e.g. see [Serre, §5.5, Proposition 38]) an exact sequence of pointed sets

$$1\to \mathbf{R}^\times\to \mathrm{Aut}(\mathbf{G}_{m,\mathbf{R}})\to \mathbf{Z}/2\mathbf{Z}\to H^1(\Gamma,\mathbf{C}^\times)\to H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))\to H^1(\Gamma, \mathbf{Z}/2\mathbf{Z}).$$

As already observed by Thomas Preu, by Hilbert's Theorem 90 (e.g. see [Poonen, Proposition 1.3.15]) that $H^1(\Gamma,\mathbf{C}^\times)$ is trivial. Moreover, as $\Gamma$ acts trivially on $\mathbf{Z}/2\mathbf{Z}$ (which is abelian) we have that

$$H^1(\Gamma,\mathbf{Z}/2\mathbf{Z})=Z^1(\Gamma,\mathbf{Z}/2\mathbf{Z})=\mathrm{Hom}(\Gamma,\mathbf{Z}/2\mathbf{Z})=\mathbf{Z}/2\mathbf{Z}.$$

Now, by the definition of exact sequence of pointed sets, the fiber over $1$ of the map

$$H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))\to H^1(\Gamma, \mathbf{Z}/2\mathbf{Z})\qquad (1)$$

is trivial, and so to understand $H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))$ it suffices to understand the fiber over the non-trivial element in the map in $(1)$.

That said, let us observe that by [Serre, §5.5, Corollary 2] and our discussion above we know that the fiber of the map in $(1)$ over the non-trivial element can be understood as

$$H^1(\Gamma, {}_\varsigma(\mathbf{C}^\times))/H^0(\Gamma,{}_\varsigma (\mathbf{Z}/2\mathbf{Z}))$$

where $\varsigma$ is any element of $Z^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))$ whose associated cocycle is non-trivial, and where $_\varsigma(-)$ is as in [Serre, §5.3]. Now, explicitly let us take $\varsigma$ corresponding $S^1$ which is given by

$$\varsigma\colon \Gamma\to \mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}})=\mathbf{C}^\times\rtimes (\mathbf{Z}/2\mathbf{Z}),\qquad \sigma\mapsto (1,-1),$$

where $\sigma$ is the complex conjugation map. We can then explicitly compute that $_\varsigma (\mathbf{C}^\times)$ is the group which has the same underlying group, but now for which we have

$$\sigma\cdot \alpha=\sigma(\alpha)^{-1}.$$

Similarly, $_\sigma(\mathbf{Z}/2\mathbf{Z})$ has the same underlying group, and now

$$\sigma\cdot -1= (-1) (-1) (-1)^{-1}=-1,$$

or, in other words, $_\varsigma(\mathbf{Z}/2\mathbf{Z})$ is still just $\mathbf{Z}/2\mathbf{Z}$ with the trivial action.

Now, as $_\varsigma(\mathbf{C}^\times)$ is abelian, and $\Gamma$ is cyclic, we can use Thomas Preu's favorite formula to compute that

$$H^1(\Gamma,{}_\varsigma(\mathbf{C}^\times))=\ker(N)/\mathrm{im}(\Delta).$$

Observe here now that

$$N\colon {}_\varsigma(\mathbf{C}^\times)\to {}_\varsigma(\mathbf{C}^\times),\qquad \alpha\mapsto \sigma(\alpha)^{-1}\alpha,$$

and so the kernel of this map is those $\alpha$ such that $$\sigma(\alpha)=\alpha$$, which is precisely $\mathbf{R}^\times$. On the other hand,

$$\Delta\colon {}_\varsigma(\mathbf{C}^\times)\to {}_\varsigma(\mathbf{C}^\times),\qquad \alpha\mapsto \sigma(\alpha)^{-1}\alpha^{-1},$$

which has image precisely $\mathbf{R}^{>0}$. So, we see that

$$H^1(\Gamma,{}_\varsigma(\mathbf{C})^\times)=\mathbf{R}^\times/\mathbf{R}^{>0}.$$

Now, the action of $_\varsigma(\mathbf{Z}/2\mathbf{Z})$ on this group is by inversion which, is clearly the trivial action, and so all-in-all we see that the fiber over the non-trivial element of the map in $(1)$ is in bijection with

$$H^1(\Gamma,{}_\varsigma(\mathbf{C})^\times)=\mathbf{R}^\times/\mathbf{R}^{>0},$$

which has $2$ elements. Thus, we see that

$$\# H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))=1+2=3.$$

Remark:

1. This was VERY long-winded because I wanted to show all the steps to illustrate the method. That said, in practice I was able to immediately compute the correct answer was 3 using this method. The utility (at least here) is that I never had to actually think about non-abelian group cohomology and, ultimately, could use Thomas Preu's favorite formula.
2. Bonus question. This computation of $H^1(\Gamma,{}_\varsigma(\mathbf{C})^\times)$ looks quite similar, at the end, to the calculation of $\mathrm{Br}(\mathbf{R})=H^2(\Gamma,\mathbf{C}^\times)$. On the other hand, the curve $Y$ has smooth compactification a Brauer--Severi variety (a twist of $\mathbf{P}^n_\mathbf{R}$ for some $n$, in this case $n=1$) which is exactly what the Brauer group computes. What is the relationship?

EDIT: Just for fun here is an answer to my second remark.

Let us observe that we have a natural map

$$\mathrm{Twist}(\mathbf{G}_{m,\mathbf{R}})\to \mathrm{Twist}(\mathbf{P}^1_{\mathbf{R}}),\qquad C\mapsto \overline{C},$$

where $\overline{C}$ is the (unique) smooth compactification of $Y$ (e.g. see Tag 0BXX). This map is not an injection though:

$$\overline{S^1}\cong \overline{\mathbf{G}_{m,\mathbf{R}}}\cong \mathbf{P}^1_{\mathbf{R}}.$$

We can model this map of sets group theoretically, and this can actually help us recompute $H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))$. Namely, we have an injection of $\Gamma$-groups (we can even upgrade this to an embedding of algebraic groups)

$$\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}})\hookrightarrow \mathrm{Aut}(\mathbf{P}^1_{\mathbf{C}})\qquad \mathbf{(2)}$$

given by sending $f$ to its extension (in the sense of Tag 0BXY) which is still an automorphism by Tag 0BY1. One can explicitly check that the following diagram commutes

$$\begin{matrix}\mathrm{Twist}(\mathbf{G}_{m,\mathbf{R}}) & \to & \mathrm{Twist}(\mathbf{P}^1_{\mathbf{R}})\\ \downarrow & & \downarrow\\ H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}})) & \to & H^1(\Gamma,\mathrm{Aut}(\mathbf{P}^1_{\mathbf{C}}))\end{matrix},$$

where the vertical maps are bijections.

We may explicitly describe the image of $\mathbf{(2)}$ as the set of automorphisms $\varphi$ of $\mathbf{P}^1_\mathbf{C}$ such that $\varphi(S)=S$ where $S=\{[1:0],[0:1]\}$. Using the identification of $\Gamma$-groups

$$\mathrm{Aut}(\mathbf{P}^1_\mathbf{C})=\mathrm{PGL}_2(\mathbf{C}),$$

where the latter acts by fractional linear transformations (e.g. see [Mumford, Chapter 0, §5]) we may then make explicitly identify the following $\Gamma$-groups

$$\begin{aligned}\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}) &= \left\{\varphi \in \mathrm{Aut}(\mathbf{P}^1_\mathbf{C}):\varphi(S)=S\right\}\\ &=\left\{\begin{pmatrix}a & 0\\ 0 & d\end{pmatrix}\right\}\cup \left\{\begin{pmatrix}0 & b\\ c & 0\end{pmatrix}\right\}\\ &= C(\gamma)\end{aligned},$$

where $\gamma=\left(\begin{smallmatrix}-1 & 0 \\ 0 & 1\end{smallmatrix}\right)$ and $C(\gamma)$ is the centerlizer of $\gamma$ in $\mathrm{PGL}_2(\mathbf{C})$. In particular, observe that we have an identification of $\Gamma$-sets

$$ \mathrm{Aut}(\mathbf{P}^1_{\mathbf{C}})/\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}})\cong \mathcal{O}_\gamma,$$

where the latter is the conjugacy class of $\gamma$ in $\mathrm{PGL}_2(\mathbf{C})$.

Now, by [Serre, §5.4, Proposition 36] and the above discussion we have exact sequence of pointed sets

$$1\to \mathrm{Aut}(\mathbf{G}_{m,\mathbf{R}})\to \mathrm{Aut}(\mathbf{P}^1_{\mathbf{R}})\to \mathcal{O}_\gamma^\Gamma\to H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))\to H^1(\Gamma,\mathrm{Aut}(\mathbf{P}^1_{\mathbf{C}})).$$

Now, with the identification of $\Gamma$-groups $\mathrm{Aut}(\mathbf{P}^1_\mathbf{C})=\mathrm{PGL}_2(\mathbf{C})$ and the short exact sequence of $\Gamma$-groups

$$1\to \mathbf{C}^\times\to \mathrm{GL}_2(\mathbf{C})\to \mathrm{PGL}_2(\mathbf{C})\to 1$$

we may use [Serre, §5.7, Proposition 43] to obtain an exact sequence of pointed sets

$$H^1(\Gamma,\mathrm{GL}_2(\mathbf{C}))\to H^1(\Gamma,\mathrm{PGL}_2(\mathbf{C}))\to H^2(\Gamma,\mathbf{C}^\times)=:\mathrm{Br}(\mathbf{R})\cong \mathbf{R}^\times/\mathbf{R}^{>0}.$$

Again by Hilbert's theorem 90 (e.g. in this context see [GS, Example 2.3.4] the first term vanishes, and in fact the map we get an injection $H^1(\Gamma,\mathrm{PGL}_2(\mathbf{C}))\to H^2(\Gamma,\mathbf{C}^\times)$ (see [GS, Theorem 4.4.5]) and in fact must be an isomorphism as the target is isomorphic to $\mathbf{Z}/2\mathbf{Z}$ and the source is non-trivial (as $\mathrm{Twist}(\mathbf{P}^1_\mathbf{R})$ contains the non-trivial element given by $\overline{Y}$). Thus, we see that $H^1(\Gamma,\mathrm{PGL}_2(\mathbf{C}))$ is a two element set.

We see that $\mathbf{G}_m$, corresponding to the trivial cocycle, and $Y$ corresponding to the cocycle

$$\vartheta\colon \Gamma\to \mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}})\subseteq \mathrm{PGL}_2(\mathbf{C}),\qquad \sigma\mapsto \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix},$$

form a set of representatives of $H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))$ which surject onto $H^1(\Gamma,\mathrm{Aut}(\mathbf{P}^1_\mathbf{C}))$. Thus, to determine the size of $H^1(\Gamma,\mathrm{PGL}_2(\mathbf{C}))$ it suffices to determine the fibers over (the images) of each of these cocycles. By [Serre, §5.4, Corollary 2] these fibers are in bijection, respectively, the following sets

$$\mathcal{O}_\gamma^\Gamma/\mathrm{PGL}_2(\mathbf{R}),\qquad \left({}_\vartheta \mathrm{PGL}_2(\mathbf{C})/{}_\vartheta C(\gamma)\right)^\Gamma/({}_\vartheta \mathrm{PGL}_2(\mathbf{C}))^\Gamma.\qquad \mathbf{(3)}$$

It is a fun exercise in linear algebra (NB: I want to thank my friend Alexander Bertoloni Meli for helping me do this exercise), to compute that

$$\mathcal{O}_\gamma=\mathrm{PGL}_2(\mathbf{R})\cdot \gamma\sqcup \mathrm{PGL}_2(\mathbf{R})\cdot \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$$

The non-triviality is a shadow of the deep notion of geometric conjugacy vs. rational conjugacy. In any case, using this it's easy to compute that the sizes of the sets in $\mathbf{(3)}$ is $2$ and $1$ respectively rederiving the calculation that $\# H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))=3$.

Finally, an answer to my second remark. From the above we have a natural map

$$H^1(\Gamma, {}_\varsigma(\mathbf{C}^\times))=H^1(\Gamma,{}_\varsigma(\mathbf{C}^\times))/H^0(\Gamma,{}_\varsigma(\mathbf{Z}/2\mathbf{Z}))\hookrightarrow H^1(\Gamma,\mathrm{Aut}(\mathbf{G}_{m,\mathbf{C}}))\to \mathrm{Br}(\mathbf{R}),$$

and this map is a bijection, so it's not very surprising that they produced 'the same computation'.

References:

[GS] Gille, P. and Szamuely, T., 2017. Central simple algebras and Galois cohomology (Vol. 165). Cambridge University Press.

[Mumford] Mumford, D., Fogarty, J. and Kirwan, F., 1994. Geometric invariant theory (Vol. 34). Springer Science & Business Media.

[Poonen] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..

[Serre] Serre, J.P., 1994. Cohomologie galoisienne (Vol. 5). Springer Science & Business Media.

Answer 2

As Alex Youcis encouraged me to do so, I try to answer myself.

The mistake was that $A:=\text{Aut}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\not\cong K^{\times}\times\mathbb{Z}/2\mathbb{Z}$ (an abelian group) but rather $\cong K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}$ (a non-abelian group). Explicitly the group composition is given by $(a_1,b_1)\cdot(a_2,b_2)=(a_1\cdot a_2^{b_1},b_1\cdot b_2)$.

Since the $N$-$\Delta$-approach only computes group cohomology when the $G$-set in question is an abelian group where the abelian group structure is compatible with the $G$-action (i.e. a $G$-module) we cannot compute $\text{H}^1_{\text{grp}}(\text{G},A)$ via $\text{ker}(N)/\text{im}(\Delta)\cong\mathbb{Z}/2\mathbb{Z}$.

The correct $\text{Tw}(\mathbb{G}_m)$ should be computed via non-abelian group cohomology as presented e.g. in J.S. Milne's lecture notes "Algebraic groups", sec. 27.a, pp. 469ff. We use 1-cocycles and equivalence $\sim$ of 1-cocycles as described there.

A 1-cocycle in this context is given by a pair of elements of $A\cong K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}$ of the form $c=(c_{\text{id}},c_{\sigma})$. The cocycle condition immediately yields $c_{\text{id}}=1_A=(1,1)$ and we will discard it from the notation and identify $c=c_{\sigma}=(a,b)=(s+ti,b)$. Exploiting the cocycle condition further we get the set of 1-cocycles to be $$\text{Cocyc}^1=\{(s+ti,1):(s,t)\in k^2\setminus\{(0,0)\} \wedge s^2+t^2-1=0\} \cup \{(s,-1):s\in k^{\times}\}$$ $$= \{(a,1):a\in\text{U }(1)\}\quad\dot{\cup}\quad\{(a,-1):a\in\mathbb{R}^+\}\quad\dot{\cup}\quad\{(a,-1):a\in\mathbb{R}^-\}.$$ Straight forward computation yields that these three disjoint sets are exactly the 3 equivalence classes of 1-cocycles.

In summary $\text{H}^1_{\text{grp}}(\text{G},A)=\left(\text{Cocyc}^1/\sim\right)=\{\mathbb{G}_m,\text{S}^1,Y\}=\text{Tw}(\mathbb{G}_m)$

Alternatively Alex Youcis suggested to use the short exact sequence of non-abelian groups (recall that abelian groups are also non-abelian groups) $$1\rightarrow K^{\times} \rightarrow K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 1,$$ which yields a long exact sequence of pointed sets, whose relevant part is given by $$\ldots\rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}) \rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}) \rightarrow \text{H}^1_{\text{grp}}(\text{G},\mathbb{Z}/2\mathbb{Z}).$$ By the classic Hilbert's Theorem 90 (cohomological version) we have $\text{H}^1_{\text{grp}}(\text{G},K^{\times})\cong 1$ and by remark 1 from the question $\text{H}^1_{\text{grp}}(\text{G},\mathbb{Z}/2\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$, which makes the previous sequence more explicit $$\ldots\rightarrow 1 \rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}) \xrightarrow{\alpha} \mathbb{Z}/2\mathbb{Z}.$$ $\text{H}^1_{\text{grp}}(\text{G},A)$ consists of the fibers above $\mathbb{Z}/2\mathbb{Z}\cong\{\mathbb{G}_m, \text{S}^1\}$ of which $\alpha^{-1}\left(\mathbb{G}_m\right)=\{\mathbb{G}_m\}$ must be a singleton, because of the properties of exact sequences of pointed sets. This leaves us with computing the other fiber $\alpha^{-1}\left(\text{S}^1\right)$.

Computing $\alpha^{-1}\left(\text{S}^1\right)$ explicitly seems to me not much simpler than computing $\text{H}^1_{\text{grp}}(\text{G},A)$ directly using 1-cocycles and equivalences as done above. At least I do not see how to do it differently, which gives this approach the feeling of a dead end. But more apt people might have better ideas than me.

Edit note: removed the trailing dots in the long exact sequences as suggested by Alex Youcis.