[proofplan] We prove the statement by reducing the complexified real exterior algebra to the standard finite-dimensional Lefschetz representation attached to a Hermitian [vector space](/page/Vector%20Space). The Lefschetz operator $L$ and its adjoint $\Lambda$ satisfy the usual $\mathfrak{sl}_2$ commutation relations, so each degree decomposes into primitive Lefschetz strings. On a primitive string generated in degree $m$, the map $L^{n-k}$ sends the summand $L^rP^m$ in degree $k$ to the symmetric summand $L^{n-m-r}P^m$ in degree $2n-k$, where the relation between the indices is \begin{align*} k=m+2r. \end{align*} The $\mathfrak{sl}_2$ string formula shows that this summand map is bijective. Taking the direct sum over all primitive strings proves that $L^{n-k}$ is an isomorphism on $E^k$. [/proofplan]

[step:Pass to the complexified exterior algebra with its Lefschetz operators] For each integer $j$, write \begin{align*} E^j=\Lambda^j V_{\mathbb R}^*\otimes_{\mathbb R}\mathbb C. \end{align*} The Hermitian metric $h$ induces a Hermitian [inner product](/page/Inner%20Product) on $V_{\mathbb R}^*\otimes_{\mathbb R}\mathbb C$ by duality and complex linear extension, and hence induces the exterior-power Hermitian inner product on each finite-dimensional complex vector space $E^j$. Let \begin{align*} \Lambda:E^j\to E^{j-2} \end{align*} denote the adjoint of $L:E^{j-2}\to E^j$ with respect to this induced inner product. Define the grading operator \begin{align*} H:E^j\to E^j \end{align*} by \begin{align*} H(\alpha)=(j-n)\alpha. \end{align*} We now verify the hypotheses of [citetheorem:8053]. The present $V$ is a Hermitian vector space of complex dimension $n$, $V_{\mathbb R}$ is its underlying real vector space, $E^j=\Lambda^jV_{\mathbb R}^*\otimes_{\mathbb R}\mathbb C$, $E=\bigoplus_{j=0}^{2n}E^j$, and $L$ is exterior multiplication by the associated Kähler form $\omega$. The operator $\Lambda$ is, by construction, the adjoint of $L$ for the Hermitian inner product induced by $h$, and $H$ acts on degree $j$ as multiplication by $j-n$. Therefore [citetheorem:8053] applies and gives \begin{align*} [\Lambda,L]=H, \end{align*} \begin{align*} [H,L]=2L, \end{align*} and \begin{align*} [H,\Lambda]=-2\Lambda. \end{align*} These are precisely the $\mathfrak{sl}_2$ relations on the finite-dimensional graded complex vector space \begin{align*} E=\bigoplus_{j=0}^{2n}E^j. \end{align*} [/step]

[step:Decompose each degree into primitive Lefschetz strings]For each integer $m$ with $0\le m\le n$, define the primitive subspace \begin{align*} P^m=\ker(\Lambda:E^m\to E^{m-2}). \end{align*} Since $E$ is finite-dimensional and $L$, $\Lambda$, and $H$ satisfy the $\mathfrak{sl}_2$ relations established above, the finite-dimensional representation theorem for $\mathfrak{sl}_2$ applies to $E$. It gives the [primitive Lefschetz decomposition](/theorems/3855) \begin{align*} E^k=\bigoplus_{r\ge 0} L^rP^{k-2r} \end{align*} for every $0\le k\le n$, where terms with $k-2r<0$ are omitted. More explicitly, for each primitive vector $p\in P^m$, the irreducible string generated by $p$ is \begin{align*} p,\quad Lp,\quad L^2p,\quad \dots,\quad L^{n-m}p, \end{align*} with $L^{n-m+1}p=0$, and the last nonzero degree in the string is $2n-m$.[/step]

[guided]The useful structure is not merely that $E$ is graded; it is that the three operators $L$, $\Lambda$, and $H$ form a finite-dimensional $\mathfrak{sl}_2$ representation. The primitive vectors are the lowest vectors for this convention: a vector $p\in E^m$ is primitive exactly when \begin{align*} \Lambda p=0. \end{align*} Starting from such a primitive vector, repeated application of $L$ gives the string \begin{align*} p,\quad Lp,\quad L^2p,\quad \dots. \end{align*} Because $E^j=\{0\}$ for $j>2n$, this string must terminate. More precisely, the finite-dimensional representation theorem for $\mathfrak{sl}_2$ applies because $E$ is finite-dimensional and the previous step established \begin{align*} [\Lambda,L]=H, \end{align*} \begin{align*} [H,L]=2L, \end{align*} and \begin{align*} [H,\Lambda]=-2\Lambda. \end{align*} In that representation, a primitive vector $p\in P^m$ is a lowest-weight vector of $H$-weight $m-n$. The irreducible $\mathfrak{sl}_2$ string generated by $p$ has highest degree $2n-m$, so its nonzero part is \begin{align*} p,\quad Lp,\quad L^2p,\quad \dots,\quad L^{n-m}p, \end{align*} and $L^{n-m+1}p=0$. Thus a primitive degree-$m$ vector contributes one summand in each degree \begin{align*} m,\quad m+2,\quad m+4,\quad \dots,\quad 2n-m. \end{align*} The same finite-dimensional $\mathfrak{sl}_2$ decomposition gives the direct sum of all such irreducible strings. Therefore every element of $E^k$ can be written uniquely as a finite sum of elements of the form $L^rp$ with \begin{align*} p\in P^{k-2r}. \end{align*} Equivalently, \begin{align*} E^k=\bigoplus_{r\ge 0}L^rP^{k-2r}. \end{align*} The directness is important: once we prove that $L^{n-k}$ is an isomorphism on each primitive-string summand, the direct sum decomposition lets us assemble those isomorphisms into an isomorphism on all of $E^k$.[/guided]

[step:Compute where $L^{n-k}$ sends one primitive summand] Fix an integer $k$ with $0\le k\le n$. Let $r\ge 0$ and set \begin{align*} m = k - 2r. \end{align*} Assume $m\ge 0$, and let $p\in P^m$. The element $L^rp$ lies in $E^k$. Applying $L^{n-k}$ gives \begin{align*} L^{n-k}(L^rp)=L^{n-k+r}p. \end{align*} Since \begin{align*} m = k - 2r, \end{align*} we obtain \begin{align*} n-k+r = n-m-r. \end{align*} Therefore \begin{align*} L^{n-k}(L^rp)=L^{n-m-r}p. \end{align*} The exponent $n-m-r$ lies between $0$ and $n-m$, because $0\le r\le k/2$ and $m\ge 0$. Hence $L^{n-m-r}p$ is a member of the same primitive string generated by $p$. The degree of this element is \begin{align*} m + 2(n-m-r)=2n-k. \end{align*} Thus $L^{n-k}$ sends the summand $L^rP^m\subset E^k$ into the summand $L^{n-m-r}P^m\subset E^{2n-k}$. [/step]

[step:Use the string formula to prove bijectivity on each primitive summand] We now prove that the map \begin{align*} L^{n-k}:L^rP^m\to L^{n-m-r}P^m \end{align*} is bijective. Since $p\mapsto L^rp$ is injective on $P^m$ for $0\le r\le n-m$, every element of $L^rP^m$ has a unique form $L^rp$ with $p\in P^m$. Under this parametrization, the map is \begin{align*} L^rp\mapsto L^{n-m-r}p. \end{align*} Again by the $\mathfrak{sl}_2$ string relation, $p\mapsto L^{n-m-r}p$ is injective on $P^m$. Therefore the displayed map is injective. It is also surjective: every element of $L^{n-m-r}P^m$ has the form $L^{n-m-r}p$ for some $p\in P^m$, and it is the image of $L^rp\in L^rP^m$. Hence $L^{n-k}$ restricts to a $\mathbb C$-linear isomorphism on each primitive summand. [/step]

[step:Assemble the primitive-string isomorphisms] The primitive decompositions in degrees $k$ and $2n-k$ are \begin{align*} E^k=\bigoplus_{r\ge 0}L^rP^{k-2r} \end{align*} and \begin{align*} E^{2n-k}=\bigoplus_{r\ge 0}L^{n-k+r}P^{k-2r}. \end{align*} The previous step shows that $L^{n-k}$ maps each summand $L^rP^{k-2r}$ isomorphically onto the corresponding summand $L^{n-k+r}P^{k-2r}$. Since both decompositions are direct sums and $L^{n-k}$ respects this summand-by-summand correspondence, $L^{n-k}:E^k\to E^{2n-k}$ is a $\mathbb C$-linear isomorphism. This proves the asserted pointwise Hard Lefschetz isomorphism for every $0\le k\le n$. [/step]