Let $X$ be a variety over $k$ and $G$ be a finite abelian group. Then we know that $H_{fppf}^{2}(X,G)$ is in bijective correspondence with isomorphism classes of $G$-banded gerbes.

Now we consider a concrete example. Let $G=\mu_{2}$ be the $2$-cyclic group and $X=\operatorname{Spec}(\mathbb{R})$. Then $H_{fppf}^{2}(X,\mu_{2})=H_{\acute{e}tale}^{2}(X,\mu_{2})\cong \mathbb{Z}/2\mathbb{Z} $ since $H_{\acute{e}tale}^{2}(X,\mu_{2})$ is just the $2$-torsion Brauer class of $\mathbb{R}$. So there are two $\mu_{2}$-gerbe over $X$. One is the classifying stack $\mathbf{B}\mu_{2,\mathbb{R}}\to \operatorname{Spec}\mathbb{R}$ which is corresponding to the trivial element. Another I denote as $\mathscr{X}$. I want to know what $\mathscr{X}$ looks like and how to describe it.

Since $H_{\acute{e}tale}^{2}(\mathbb{C},\mu_{2})$ is trivial, then we have the following commutative diagram \begin{array}{ccc} \mathbf{B}\mu_{2,\mathbb{C}} & \xrightarrow{} & \mathscr{X} \\ \downarrow & & \downarrow \\ \mathrm{Spec}\mathbb{C} & \xrightarrow{} & \mathrm{Spec}\mathbb{R} \end{array}.

At the beginning, I thought maybe $\mathscr{X}=[\mathrm{Spec}\mathbb{R}[x]/(x^{2}+1)/\mu_{2}]$, where $\zeta\cdot x=-x$. Here $\zeta$ is a generator of $\mu_{2}$. But in this case it seems $[\mathrm{Spec}\mathbb{R}[x]/(x^{2}+1)/\mu_{2}]\otimes_{\mathbb{R}}\mathbb{C}\cong [\mathrm{Spec}\mathbb{C}[x]/(x^{2}+1)/\mu_{2}]\cong \mathrm{Spec}\mathbb{C} $ since now the action is free. So it is not correct.

$\mathscr{X}$ should be a quotient stack. I wonder how to describe it as a quotient stack or describe it as category.

Let $X$ be a variety over $k$ and $G$ be a finite abelian group. Then we know that $H_{fppf}^{2}(X,G)$ is in bijective correspondence with isomorphism classes of $G$-banded gerbes.

Now we consider a concrete example. Let $G=\mu_{2}$ be the $2$-cyclic group and $X=\operatorname{Spec}(\mathbb{R})$. Then $H_{fppf}^{2}(X,\mu_{2})=H_{\acute{e}tale}^{2}(X,\mu_{2})\cong \mathbb{Z}/2\mathbb{Z} $ since $H_{\acute{e}tale}^{2}(X,\mu_{2})$ is just the $2$-torsion Brauer class of $\mathbb{R}$. So there are two $\mu_{2}$-gerbe over $X$. One is the classifying stack $\mathbf{B}\mu_{2,\mathbb{R}}\to \operatorname{Spec}\mathbb{R}$ which is corresponding to the trivial element. Another I denote as $\mathscr{X}$. I want to know what $\mathscr{X}$ looks like and how to describe it.

Since $H_{\acute{e}tale}^{2}(\mathbb{C},\mu_{2})$ is trivial, then we have the following commutative diagram \begin{array}{ccc} \mathbf{B}\mu_{2,\mathbb{C}} & \xrightarrow{} & \mathscr{X} \\ \downarrow & & \downarrow \\ \mathrm{Spec}\mathbb{C} & \xrightarrow{} & \mathrm{Spec}\mathbb{R} \end{array}.

At the beginning, I thought maybe $\mathscr{X}=[\mathrm{Spec}\mathbb{R}[x]/(x^{2}+1)/\mu_{2}]$, where $\zeta\cdot x=-x$. Here $\zeta$ is a generator of $\mu_{2}$. But in this case it seems $[\mathrm{Spec}\mathbb{R}[x]/(x^{2}+1)/\mu_{2}]\otimes_{\mathbb{R}}\mathbb{C}\cong [\mathrm{Spec}\mathbb{C}[x]/(x^{2}+1)/\mu_{2}]\cong \mathrm{Spec}\mathbb{C} $ since now the action is free. So it is not correct.

Now I think maybe $\mathscr{X}\cong \mathbf{B}\mu_{2,\mathbb{R}}\times[\mathrm{Spec}\mathbb{C}[x]/(x^{2}+1)/\mu_{2}]$. Am I correct?