[proofplan] We describe tangent vectors as first-order deformations over the dual numbers $\mathbb C[\varepsilon]/(\varepsilon^2)$. Every such first-order deformation is obtained by applying an operator $1+\varepsilon A$ to the fixed filtration, and two operators give the same first-order deformation precisely when they differ by an endomorphism preserving every member of the filtration. This gives the canonical quotient description. A choice of splitting then decomposes this quotient into the displayed direct sum of homomorphism spaces, and differentiating the polarization equations gives the period-domain tangent subspace. [/proofplan]

[step:Model first-order deformations by the dual-number filtration]Let \begin{align*} R:=\mathbb C[\varepsilon]/(\varepsilon^2) \end{align*} be the algebra of dual numbers, and let \begin{align*} H_R:=H\otimes_{\mathbb C}R \end{align*} be the corresponding free $R$-module. A tangent vector to $\operatorname{Flag}(H,(\dim_{\mathbb C}F^p)_p)$ at $F^\bullet$ is represented by a decreasing filtration $\widetilde F^\bullet$ of $H_R$ by free direct-summand $R$-submodules such that \begin{align*} \widetilde F^p\otimes_R R/(\varepsilon)=F^p \end{align*} for every $p$, such that each $\widetilde F^p$ has the same $R$-rank as $F^p$ has complex dimension, and such that each subquotient $\widetilde F^p/\widetilde F^{p+1}$ is locally free of the prescribed rank. Since $R$ is the local Artinian ring $\mathbb C[\varepsilon]/(\varepsilon^2)$, locally free $R$-modules are free. Therefore $\widetilde F^p/\widetilde F^{p+1}$ is a free $R$-module whose reduction modulo $\varepsilon$ is $F^p/F^{p+1}$. Choose, for each $p$, a complementary complex subspace $G^p\subset F^p$ to $F^{p+1}$, so that \begin{align*} F^p=G^p\oplus F^{p+1}. \end{align*} Iterating this decomposition gives a direct-sum splitting \begin{align*} H=\bigoplus_p G^p \end{align*} with \begin{align*} F^p=\bigoplus_{r\ge p}G^r. \end{align*} For each $p$, choose a complex basis $\mathcal B_p$ of $G^p$. We choose lifts of these basis vectors downward in the filtration. Suppose lifts have already been chosen for all $\mathcal B_r$ with $r>p$ and form an $R$-basis of $\widetilde F^{p+1}$. By the quotient-freeness just established, $\widetilde F^p/\widetilde F^{p+1}$ is a free $R$-module reducing to $G^p$ modulo $\varepsilon$. Choose, for each $g\in\mathcal B_p$, an element $\widetilde g\in\widetilde F^p$ whose image modulo $\varepsilon$ is $g$ and whose classes form an $R$-basis of $\widetilde F^p/\widetilde F^{p+1}$. Then the previously chosen lifts together with these $\widetilde g$ form an $R$-basis of $\widetilde F^p$ by [Nakayama's lemma](/theorems/2935) for the local Artinian ring $R$. Each chosen lift has the form $\widetilde g=g+\varepsilon u_g$ with $u_g\in H$. Extending the assignment $g\mapsto u_g$ complex-linearly over the direct-sum basis $\bigcup_p\mathcal B_p$ defines a complex-[linear map](/page/Linear%20Map) \begin{align*} A:H\to H. \end{align*} By the basis construction, for every $p$ the $R$-basis of $\widetilde F^p$ is exactly the image under $1+\varepsilon A$ of the corresponding basis of $F^p\otimes_{\mathbb C}R$. Hence \begin{align*} \widetilde F^p=(1+\varepsilon A)(F^p\otimes_{\mathbb C}R) \end{align*} inside $H_R$ for every $p$.[/step]

[guided]The point of using the dual numbers is that an element of $H_R$ has the form $x+\varepsilon y$, with $x,y\in H$, and the relation $\varepsilon^2=0$ records only first-order motion. A first-order deformation of the flag is therefore a filtration $\widetilde F^\bullet$ in $H_R$ whose reduction at $\varepsilon=0$ is the original filtration $F^\bullet$. In the flag variety, these are filtrations by free direct-summand $R$-submodules with the prescribed ranks and with locally free subquotients of the prescribed ranks. Since $R=\mathbb C[\varepsilon]/(\varepsilon^2)$ is local Artinian, every locally free $R$-module is free. Therefore each quotient $\widetilde F^p/\widetilde F^{p+1}$ is free over $R$ and reduces modulo $\varepsilon$ to $F^p/F^{p+1}$. We choose subspaces $G^p\subset F^p$ satisfying \begin{align*} F^p=G^p\oplus F^{p+1}. \end{align*} This is a splitting of the filtration as a complex [vector space](/page/Vector%20Space). Repeating the decomposition gives \begin{align*} H=\bigoplus_pG^p, \end{align*} and the filtration is recovered from the splitting by \begin{align*} F^p=\bigoplus_{r\ge p}G^r. \end{align*} For each $p$, choose a complex basis $\mathcal B_p$ of $G^p$. We now choose lifts in descending order of $p$. Assume that the basis vectors in all $G^r$ with $r>p$ have already been lifted and that those lifts form an $R$-basis of $\widetilde F^{p+1}$. The quotient $\widetilde F^p/\widetilde F^{p+1}$ is a free $R$-module whose reduction modulo $\varepsilon$ is $G^p$, by the direct-summand condition in the first-order flag functor. Hence we may choose lifts $\widetilde g\in\widetilde F^p$ of the basis vectors $g\in\mathcal B_p$ whose classes form an $R$-basis of that quotient. By Nakayama's lemma for the local Artinian ring $R=\mathbb C[\varepsilon]/(\varepsilon^2)$, these new lifts together with the already chosen lifts form an $R$-basis of $\widetilde F^p$. Each lift has the unique form $\widetilde g=g+\varepsilon u_g$ with $u_g\in H$. Defining $A(g):=u_g$ on every basis vector $g\in\bigcup_p\mathcal B_p$ and extending complex-linearly gives a map \begin{align*} A:H\to H. \end{align*} Because the chosen lifted basis of $\widetilde F^p$ is exactly the image of the basis of $F^p\otimes_{\mathbb C}R$ under $1+\varepsilon A$, the deformed filtration satisfies \begin{align*} \widetilde F^p=(1+\varepsilon A)(F^p\otimes_{\mathbb C}R). \end{align*} This proves that every tangent vector can be represented by an endomorphism $A$ of $H$.[/guided]

[step:Identify the endomorphisms that give the zero tangent vector] Let \begin{align*} \mathfrak p_F:=\{B\in\operatorname{End}_{\mathbb C}(H):B(F^p)\subseteq F^p\text{ for every }p\} \end{align*} be the [Lie algebra](/page/Lie%20Algebra) of the stabilizer of the filtration $F^\bullet$. If $B\in\mathfrak p_F$, then \begin{align*} (1+\varepsilon B)(F^p\otimes_{\mathbb C}R)=F^p\otimes_{\mathbb C}R \end{align*} for every $p$, so $B$ represents the zero tangent vector. Conversely, suppose that $A\in\operatorname{End}_{\mathbb C}(H)$ represents the zero tangent vector. Then \begin{align*} (1+\varepsilon A)(F^p\otimes_{\mathbb C}R)=F^p\otimes_{\mathbb C}R \end{align*} for every $p$. For each $x\in F^p$, the element \begin{align*} x+\varepsilon A(x) \end{align*} belongs to $F^p\otimes_{\mathbb C}R$. This is possible only if $A(x)\in F^p$. Hence $A(F^p)\subseteq F^p$ for every $p$, so $A\in\mathfrak p_F$. Therefore \begin{align*} T_{F^\bullet}\operatorname{Flag}(H,(\dim_{\mathbb C}F^p)_p)\cong \operatorname{End}_{\mathbb C}(H)/\mathfrak p_F. \end{align*} [/step]

[step:Decompose the quotient after choosing a splitting of the filtration] Fix a splitting \begin{align*} H=\bigoplus_pG^p \end{align*} with $G^p\cong \operatorname{Gr}_F^pH$ and \begin{align*} F^p=\bigoplus_{r\ge p}G^r. \end{align*} For an endomorphism $A\in\operatorname{End}_{\mathbb C}(H)$ and indices $p,r$, let \begin{align*} A_{rp}:G^p&\to G^r \end{align*} denote the component obtained by restricting $A$ to $G^p$ and projecting to $G^r$ along the direct sum decomposition. The condition $A(F^p)\subseteq F^p$ for every $p$ is equivalent to \begin{align*} A_{rp}=0\quad\text{whenever }r<p. \end{align*} Thus the quotient by $\mathfrak p_F$ retains exactly the components \begin{align*} A_{rp}:G^p\to G^r\quad\text{with }r<p. \end{align*} For fixed $p$, the direct sum of the targets with $r<p$ is \begin{align*} \bigoplus_{r<p}G^r\cong H/F^p. \end{align*} Hence the chosen splitting gives \begin{align*} \operatorname{End}_{\mathbb C}(H)/\mathfrak p_F \cong \bigoplus_p\operatorname{Hom}_{\mathbb C}(G^p,H/F^p). \end{align*} Using the splitting isomorphism $G^p\cong \operatorname{Gr}_F^pH$, this becomes \begin{align*} T_{F^\bullet}\operatorname{Flag}(H,(\dim_{\mathbb C}F^p)_p)\cong \bigoplus_p\operatorname{Hom}_{\mathbb C}(\operatorname{Gr}_F^pH,H/F^p). \end{align*} [/step]

[step:Differentiate the polarization equations to obtain the period-domain tangent space] Assume now that the flags are required to satisfy polarization relations defined by a [bilinear form](/page/Bilinear%20Form) \begin{align*} Q:H\times H&\to \mathbb C. \end{align*} These relations are algebraic equations on the flag variety. Let \begin{align*} \operatorname{Per}\subseteq \operatorname{Flag}(H,(\dim_{\mathbb C}F^p)_p) \end{align*} denote the corresponding locally closed locus at the fixed Hodge numbers, and let $F^\bullet\in\operatorname{Per}$. We use the standard Zariski tangent-space principle for a locally closed subvariety cut out near a point by equations: its Zariski tangent space is the common kernel of the differentials of the defining equations inside the tangent space of the ambient variety. If $F^\bullet$ lies in the smooth locus of $\operatorname{Per}$, this Zariski tangent space is the ordinary complex tangent space. A first-order deformation represented by $A\in\operatorname{End}_{\mathbb C}(H)$ is tangent to $\operatorname{Per}$ exactly when the deformed filtration \begin{align*} (1+\varepsilon A)F^\bullet \end{align*} satisfies the polarization equations modulo $\varepsilon^2$. Expanding each polarization equation and keeping the coefficient of $\varepsilon$ gives a homogeneous linear condition on the class of $A$ in \begin{align*} \operatorname{End}_{\mathbb C}(H)/\mathfrak p_F. \end{align*} Therefore \begin{align*} T_{F^\bullet}\operatorname{Per} \subseteq T_{F^\bullet}\operatorname{Flag} \end{align*} is precisely the linear subspace cut out by the infinitesimal polarization conditions. This is the asserted description of the period-domain tangent space. [/step]