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For the defining representation of the special linear Lie algebra ${\mathfrak{sl}}_3$, the crystal basis is $1\stackrel{1}{\to }2\stackrel{2}{\to }3$, with weights ${\epsilon }_i$ restricted to trace-zero diagonal matrices. Its dual is $\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights $-{\epsilon }_i$. Using the crystal tensor-product rule with lowering on the first factor when ${\phi }_i(a)>{\epsilon }_i(b)$, the tensor square has highest vertices $1\otimes 1$ and $1\otimes 2$, of highest weights $2{\omega }_1,{\omega }_2$, generating components of six and three vertices. The defining tensor dual has highest vertices $1\otimes \bar{3}$ and $1\otimes \bar{1}$, generating components of eight and one vertices and weights ${\omega }_1+{\omega }_2,0$. These encode the symmetric square plus exterior square, and the trace-free endomorphisms plus scalar identity, respectively. A crystal vertex labelled by a tensor pair need not represent that same undeformed ordinary simple tensor; the actual invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$.

In a field of characteristic two, $1+1=0$. The usual splitting of a tensor product into its symmetric square and exterior square by $(I\pm \tau )/2$ is unavailable.

The defining representation of ${\mathfrak{so}}_{2n+1}$ has weights ${\epsilon }_1,\dots ,{\epsilon }_n,0,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Its crystal basis is a chain on these vertices, with lowering-edge colors $1,\dots ,n-1,n,n,n-1,\dots ,1$. The short-root string has length two through zero. With the crystal tensor-product rule, for $n\ge 2$ the tensor square has highest vertices $1\otimes 1$, $1\otimes 2$ and $1\otimes \bar{1}$, corresponding to the traceless symmetric square, the exterior square and the trivial Lie algebra representation. Their dimensions are $(2n+1)(n+1)-1$, $n(2n+1)$ and one.

For $V=L({\omega }_1)$ with the G2 root system numbered short-first,In the chain basis of the crystal of the seven-dimensional G2 representation, normalize $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero highest-weight vectors of weights ${\omega }_2$ and ${\omega }_1$; the identities $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$, $e_2{v}_2=v_1$ verify both raising conditions. The Weyl complete reducibility theorem and G2 dimension polynomial exhaust the twenty-one dimensions of the exterior square. In the symmetric square, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a Lie-invariant bilinear form, which is symmetric because a nondegenerate alternating bilinear form cannot have odd dimension; its inverse gives the remaining invariant line.

Fix a volume form on a four-dimensional real symplectic vector space. Wedge pairing on its exterior square has signature $(3,3)$. The symplectic group fixes the bivector inverse to the symplectic form. Its orthogonal complement is five-dimensional with signature $(3,2)$ after choosing the appropriate overall sign. The induced representation has kernel $\{\pm I\}$: identity action on the entire exterior square preserves every two-plane and thus every line, forcing a scalar transformation with square one. Its derivative is injective and both group dimensions are ten, so its image is the identity component of $SO(3,2)$.

The exterior square of a direct sum splits into wedges of two vectors from $V$, two from $W$, or one from each. The mixed map sends $v\otimes w$ to $(v,0)\wedge (0,w)$. A combined basis proves it is an isomorphism, and the exterior-power Lie algebra representation proves equivariance for modules. No division by $2$ is used.

For a complex vector space with a nondegenerate symmetric bilinear form $Q$, the map $u\wedge v\mapsto [w\mapsto Q(v,w)u-Q(u,w)v]$ identifies the exterior square of the defining representation with the Adjoint representation of a Lie algebra of $\mathfrak{so}(V)$. Equivariance follows from invariance of $Q$; an orthogonal basis verifies bijectivity. It is irreducible when $\mathfrak{so}(V)$ is simple. In dimension four it splits into the adjoint modules of the two commuting sl2 Lie algebras.

On the six-dimensional exterior square of ${\mathbb{F}}_2^4$, the Pfaffian $Q=x_{12}{x}_{34}+x_{13}{x}_{24}+x_{14}{x}_{23}$ is a nonsingular plus-type quadratic form. The action of $G{L}_4(2)$ preserves it and is faithful: fixing every decomposable exterior line fixes every two-space and then every one-space. The even-weight binary subset module on eight letters, modulo the all-one vector, gives another six-dimensional plus-type space with $Q=\mathrm{wt}(x)/2\mathrm{mod}2$. Coordinate permutations embed $S_8$ faithfully into its orthogonal group. That orthogonal group has order $40320$, while $G{L}_4(2)$ has order $20160$, identifying its index-two image with the alternating group $A_8$.

For a real inner product space $V$, the displayed map identifies ${\Lambda }^{2}V$ with its Special orthogonal Lie algebra. It sends a wedge of distinct orthonormal basis vectors to the corresponding elementary skew-symmetric matrix, so it is an isomorphism. For an orthogonal transformation $g$, invariance of the inner product gives $\Phi (gu\wedge gv)=g\Phi (u\wedge v)g^{-1}$. Thus the exterior square of the defining representation is the Adjoint representation of the orthogonal group. Complexification identifies their weight spaces, allowing the root system to be read from the exterior-square weights and the zero-weight multiplicity.

Each level-two chiral sector of the critical closed string is the symmetric traceless square $S_0^{2}V$ of the massive little group vector space. The tensor product of its two sectors contains the exterior square ${\Lambda }^{2}V$: the equivariant map $(S,T)\mapsto ST-TS$ is onto the antisymmetric matrices. Compact-group complete reducibility realizes this quotient as a subrepresentation. It describes a massive two-form field with $M^2=8\pi T=4/{\alpha }^{\prime }$.

An antisymmetric tensor $A_{mn}$ with $F=dA$ obeys ${\partial }^m{F}_{mnp}={\mu }^2{A}_{np}$ for $\mu \ne 0$. Taking a divergence yields ${\partial }^m{A}_{mn}=0$, and the remaining equation is the Klein-Gordon equation for each component. The physical particle polarizations form the exterior square of the massive little group vector representation, of dimension $(D-1)(D-2)/2$. Unlike a two-form gauge field, it has physical longitudinal polarizations.

The unprojected level-one states form a transverse vector representation and an exterior square. Fermionic anticommutation makes the two-fermion states antisymmetric. Together they restrict from an antisymmetric rank-two tensor of $SO(D-1)$, not a massive spin-two field.

Take the diagonal Cartan subalgebra of ${\mathfrak{sl}}_3$ and let ${\epsilon }_i$ select its $i$th diagonal entry, so ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3=0$ on it. Use simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2=-{\epsilon }_3={\epsilon }_1+{\epsilon }_2$. The defining representation $V$ has highest vector $e_1$ and the dual has highest vector $e_3^{\ast }$. ThusTheir crystal basis diagrams are $B:1\stackrel{1}{\to }2\stackrel{2}{\to }3$ and $B^{\vee }:\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights ${\epsilon }_i$ and $-{\epsilon }_i$ respectively. Use the same crystal tensor-product rule as in2(d): lowering acts on the first factor when ${\phi }_i(u)>{\epsilon }_i(v)$; otherwise on the second. Raising uses the first factor when ${\phi }_i(u)\ge {\epsilon }_i(v)$, so it reverses each lowering arrow. An edge is absent if the selected factor operation is zero.

For $B\otimes B$, abbreviate $i\otimes j$ by $ij$. The complete color-one edge list is $11\to 21\to 22$, $31\to 32$, $13\to 23$; the complete color-two list is $21\to 31$, $22\to 32\to 33$, $12\to 13$. There are two connected components:For example ${\tilde{f}}_1(1\otimes 1)=2\otimes 1$ because ${\phi }_1(1)=1>{\epsilon }_1(1)=0$, whereas ${\tilde{f}}_1(1\otimes 2)=0$ because equality selects the second factor and ${\tilde{f}}_1(2)=0$. This explains which ordered tensor labels lie in each component. At the representation level, the two components are the six-dimensional symmetric square and the three-dimensional exterior square:

For $B\otimes B^{\vee }$, abbreviate $i\otimes \bar{j}$ by $i\bar{j}$. The complete color-one edge list is $1\bar{3}\to 2\bar{3}$, $1\bar{2}\to 2\bar{2}\to 2\bar{1}$, $3\bar{2}\to 3\bar{1}$; the complete color-two list is $1\bar{3}\to 1\bar{2}$, $2\bar{3}\to 3\bar{3}\to 3\bar{2}$, $2\bar{1}\to 3\bar{1}$. The isolated vertex is $1\bar{1}$. Thus the two highest vertices are $1\bar{3}$ of weight ${\epsilon }_1-{\epsilon }_3={\omega }_1+{\omega }_2$, and $1\bar{1}$ of weight zero. Lowering from the former visits all eight other vertices. ConsequentlyThis also follows concretely from $V\otimes V^{\ast }\cong \mathrm{End}(V)={\mathfrak{sl}}_3\oplus \mathbb{C}I$: the first term is the Adjoint representation, with highest weight the highest root, and the second is the trivial Lie algebra representation. The ordinary invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$, not the bare tensor suggested by the singleton's crystal label. The explicit edge lists and the diagrams exhibit all eighteen tensor vertices and justify the four highest weights.

Let $v_{1,+}=e_1+i{e}_2$, $v_{1,-}=e_1-i{e}_2$, $v_{2,+}=e_3+i{e}_4$, $v_{2,-}=e_3-i{e}_4$ and $v_0=e_5$. Direct multiplication by the printed rotation blocks gives their torus eigenvaluesThus the differentiated weights of a representation are $i{x}_1,-i{x}_1,i{x}_2,-i{x}_2,0$, with this sign convention fixed by the given matrix blocks.

A wedge of two distinct eigenvectors has the product eigenvalue, so its differentiated weight is the sum. The ten vectors in a basis of the exterior square split into three groups:The eight nonzero weights are distinct, so each has multiplicity one; the zero weight space has dimension two. The multiplicities add to $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, accounting for the entire representation.

Choose positive rootsThe corresponding simple roots are ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$. The vector $v_{1,+}\wedge v_{2,+}$ has weight $\lambda ={\epsilon }_1+{\epsilon }_2$. Adding any positive root to this weight produces a weight absent from part (a), so every positive root vector annihilates it. It is consequently a highest-weight vector with highest weight $\lambda$.

A representation of a compact Lie group is completely reducible: Haar averaging makes an inner product invariant, and orthogonal complements of invariant subspaces remain invariant. Therefore the exterior square contains an irreducible summand $L_{\lambda }$ of this highest weight. This can also be seen by projecting the displayed highest-weight vector into irreducible summands; a nonzero projection remains annihilated by all positive root vectors and has the same weight. Its multiplicity is one because the weight space at $\lambda$ is one-dimensional.

Use the Weyl dimension formula, expressly allowed in this part:Here $L_{\lambda }$ is the irreducible highest-weight representation, ${\Phi }^{+}$ is the chosen positive root system, $\rho$ is half its sum, and the pairing is the invariant Euclidean inner product on the real weight plane, with ${\epsilon }_1,{\epsilon }_2$ orthonormal. Using coroots instead gives the same ratio because the length factor cancels for each root. In this convention,The four factors, in the listed positive-root order, are respectivelyTheir product is $10$. Since $\dim {\Lambda }^2{\mathbb{C}}^5=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, this irreducible summand occupies the entire exterior square. HenceThe proof uses the dimension formula only where it is allowed, not in the symmetric-power argument of Question 2. Its B2 adjoint representation weight diagram is exactly the eight roots and the double zero weight obtained above.

The isomorphic pair is $A_8\cong PS{L}_4(2)$; $PS{L}_3(4)$ is different. All three have order $20160$, so order alone does not settle the question; explicitly $|PS{L}_3(4)|=(64-1)(64-4)(64-16)/(3\cdot 3)=20160$. We give a concrete realization and an element-order distinction.

On $W={\bigwedge }^2{\mathbb{F}}_2^4$, the Pfaffian isIt is a nonsingular plus-type quadratic form: the three displayed pairs give a hyperbolic decomposition. Change of basis multiplies the Pfaffian by the determinant, which is always one in $G{L}_4(2)$. Hence the exterior-square realization of PSL4 over F2 embeds $G{L}_4(2)$ into $O_6^{+}(2)$. The kernel of a group homomorphism is trivial: an element acting trivially on the exterior square fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over ${\mathbb{F}}_2$ this scalar is one.

For a second model let $E$ be the even-weight vector subspace of ${\mathbb{F}}_2^8$ and quotient by the all-one vector. The dot product descends to a nondegenerate alternating bilinear form on the six-dimensional quotient, andis well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate permutations give a faithful $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity permutation.

From the independently derived symplectic order and quadratic orbit count below,Thus $O_6^{+}(2)\cong S_8$ and the image of $G{L}_4(2)=S{L}_4(2)=PS{L}_4(2)$ is its index-two subgroup $A_8$. The index-two subgroup is unique: a nontrivial group homomorphism $S_8\to C_2$ sends every transposition to the same nonidentity element and is the sign map.

To separate $PS{L}_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PS{L}_3(4)$ lifts to a semisimple linear operator, since its order upstairs divides three times an odd order in characteristic two. The possible eigenspace degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all eigenvalues have order dividing three, so the projective element has no order 15. In the $2+1$ case the eigenvalues are $\lambda ,{\lambda }^4,{\lambda }^{-5}$ with $\lambda \in {\mathbb{F}}_{16}^{\times }$; its fifth power is scalar because ${\lambda }^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar subgroup of order three gives projective order dividing seven. There is therefore no element of order 15 in $PS{L}_3(4)$.

Several small-parameter coincidences connect the finite classical groups to permutation groups. The projective-line action immediately gives $PS{L}_2(2)\cong S_3$, $PS{L}_2(3)\cong A_4$, $PG{L}_2(3)\cong S_4$ and $PS{L}_2(4)\cong A_5$ by faithful actions of degrees three, four and five and their orders. Also $PS{L}_2(5)\cong A_5$: its involutions have Klein-four centralizers, giving five Sylow subgroups; their conjugation action is faithful and embeds its order-60 simple group into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PG{L}_2(5)$, with trivial kernel of a group homomorphism and order 120, giving $S_5$. Indeed its kernel intersects $PS{L}_2(5)$ trivially, so centralizes that normal subgroup. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.

The less immediate coincidence $PS{L}_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid 456$ and $0$ to $126\mid 345$, and label the other partitions by applying $a^x{b}^y$ to the zero partition, with $t=x+iy\in {\mathbb{F}}_9$ and $i^2=-1$. Inspection of these nine partitions givesTranslations and inversion generate $PS{L}_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $S{L}_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition permutations come from even permutations of six letters, whose partition action is faithful as proved below; hence they fill $A_6$.

Finally, on the even-weight module of ${\mathbb{F}}_2^6$ modulo its all-one vector, coordinate permutations preserve a nondegenerate alternating bilinear form in dimension four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $S{p}_4(2)$ have order 720, proving $S{p}_4(2)\cong S_6$, with commutator subgroup $A_6$. These examples illustrate both the need to quotient group centers and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for group isomorphism, as the two order-20160 examples already demonstrate.

Let $V={\mathbb{R}}^4$ have symplectic basis $(e_1,f_1,e_2,f_2)$. Fix $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$ and define a symmetric bilinear form $Q$ on ${\Lambda }^{2}V$ by $u\wedge v=Q(u,v)\mathrm{vol}$. The pairings between complementary basis bivectors split this six-dimensional space into three hyperbolic planes, so $Q$ has signature $(3,3)$.

The symplectic group preserves volume and the bivector $\Omega =e_1\wedge f_1+e_2\wedge f_2$, obtained from the inverse symplectic form, up to its fixed sign convention. Since $Q(\Omega ,\Omega )=2$, its orthogonal complement $W={\Omega }^{\perp }$ has dimension five and signature $(2,3)$. On $W$ use $-Q$, which has signature $(3,2)$. This constructs the exterior-square double cover of SO0(3,2).

If an element acts trivially on $W$, it also fixes $\Omega$, so its action on the entire exterior square is the identity. Every decomposable bivector then shows that its two-plane is preserved. A line is the intersection of two such planes; hence every line is preserved, and the matrix is scalar. Its exterior-square action forces its scalar square to equal one, giving precisely $\pm I$.

The derivative is injective by the same infinitesimal argument: zero action on ${\Lambda }^{2}V$ forces an infinitesimal scalar, whose induced scalar there is twice its value and therefore zero. The symplectic Lie algebra has dimension $2^2+2(2\cdot 3/2)=10$, from blocks $\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with symmetric $B,C$. The orthogonal Lie algebra in five dimensions also has dimension ten. Thus the image is open. Real symplectic polar decomposition has connected compact factor $U(2)$ and a contractible positive symplectic factor, so $Sp(4,\mathbb{R})$ is connected and its image is exactly the identity component:As in the preceding noncompact cases, this does not identify the quotient with the disconnected full $SO(3,2)$.

Realize the defining sl2 Lie algebra Lie algebra representation as $\mathbb{C}x\oplus \mathbb{C}y$, with $H=x{\partial }_x-y{\partial }_y$, $E=x{\partial }_y$, $F=y{\partial }_x$. A basis of the symmetric power is $e_i=x^{4-i}{y}^i$ for $0\le i\le 4$, with $H{e}_i=(4-2i)e_i$. Hence the exterior square basis $w_{ij}=e_i\wedge e_j$, $i<j$, has weight $8-2(i+j)$.

Collecting equal weights gives the complete weight-space decomposition:Every one of the ten exterior basis vectors appears exactly once. Thus $U={⨁}_{m=6,4,2,0,-2,-4,-6}{U}_m$, with the displayed bases and dimensions.

The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.