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Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.

Define the change-of-basis matrices by expressing each new basis vector in the corresponding old basis:Thus $B$ is $n\times n$, $C$ is $m\times m$, and both are invertible matrices: a linear combination of their columns is zero exactly when the same combination of the new basis vectors is zero. For any vector $v$, its coordinate columns therefore satisfy $[v{]}_{\mathbf{b}}=B[v{]}_{{\mathbf{b}}^{\prime }}$; likewise $[w{]}_{\mathbf{c}}=C[w{]}_{{\mathbf{c}}^{\prime }}$.

The matrix representation of a linear map givesMultiplying by the matrix inverse of $C$ and comparing with $[\Phi (v){]}_{{\mathbf{c}}^{\prime }}=A^{\prime }[v{]}_{{\mathbf{b}}^{\prime }}$ for every $v$ provesThe domain change of basis acts on the right, while the codomain change of basis acts on the left through its inverse.

An $r$-uniform hypergraph is strongly saturated with respect to a complete uniform hypergraph of order $r+t$ if adding any missing edge creates a new complete uniform hypergraph on $r+t$ vertices containing that edge. Equivalently, for every missing $r$-set $e$, there is an $(r+t)$-set $T\supseteq e$ for which all other $r$-subsets are edges. Existing complete subhypergraphs are allowed in this strong-saturation definition.

We prove the bound by an exterior-power quotient bound for strong hypergraph saturation. Assume $r,t\ge 1$ and $n\ge r+t$. Let $V={\mathbb{R}}^n$ with basis $e_1,\dots ,e_n$. Choose distinct real numbers $a_1,\dots ,a_n$ and let $U$ be the kernel of the $t\times n$ matrix whose column $i$ isEvery $t$ columns are linearly independent by the Vandermonde determinant, so $\dim U=n-t$. Form the quotient vector spaceFor each $r$-set $e=\{i_1<\cdots <i_r\}$, let $v_e$ be the image in $W$ of the exterior product $e_{i_1}\wedge \cdots \wedge e_{i_r}$. These images span $W$.

For an $(r+t)$-set $T$, let $V_T$ be its coordinate subspace. The matrix restricted to $T$ has rank $t$, so $U\cap V_T$ has dimension $r$. The wedge of a basis of this intersection is a nonzero vectorEvery coefficient $c_e$ is nonzero. Indeed, projection from $U\cap V_T$ onto the $r$ coordinates in $e$ is an isomorphism: a vector in its kernel would be supported on $T\setminus e$, where the $t$ columns are independent, and hence would be zero. That projection's determinant is precisely the corresponding wedge coefficient. Thus, in the quotient vector space,For a missing edge, choose its strong-saturation witness $T$. All other vectors in this relation come from present edges, so the missing edge's vector is in their span. Consequently the present-edge vectors already span all of $W$. Their number is at least its dimension:This is sharp. Take all $r$-sets that meet a fixed $t$-set $D$. A missing edge $e$ is disjoint from $D$, and on $e\cup D$ every other $r$-set meets $D$; adding $e$ completes the required clique. The number of edges is exactly the boxed difference. If $r\le n<r+t$, no missing edge could have a witness, so a strongly saturated hypergraph must be complete; the substantive construction above is for the usual range $n\ge r+t$.

For the set-pair problem, write $m=|I|$ and choose $x_i\in R_i\cap S_i$. If $i\ne j$, the cross-disjointness implies that $x_i$ is in neither $R_j$ nor $S_j$. In particular, the witnesses are distinct. Let $D=\{x_i:i\in I\}$; each $R_i$ and $S_i$ contains exactly its own witness from $D$.

Choose two different indices $i,j$. The sets $R_i\setminus \{x_i\}$ and $S_j\setminus \{x_j\}$ are disjoint, lie outside $D$, and have sizes $r-1$ and $s-1$. HenceFor equality, partition the ground set into disjoint sets $P,Q,D$ with sizes $r-1,s-1,n-r-s+2$, respectively, and enumerate $D$ as $x_1,\dots ,x_m$. SetThen $R_i\cap S_j$ is $\{x_i\}$ for $i=j$ and empty otherwise. This realizes equality whenever the bound permits $m\ge 2$. The argument is a diagonal-intersection set-pair bound, with no additional assumptions on intersections among the $R_i$ or among the $S_i$.