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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.

In the irreducible sl2 Lie algebra module $V(m)$, use $v_j=f^j{v}_0$, with $0\le j\le m$. The Lie-invariant bilinear form $B(v_j,v_k)=(-1{)}^j{1}_{\{j+k=m\}}$ is nondegenerate: its anti-diagonal is nonzero. Invariance follows from $h{v}_j=(m-2j)v_j$, $f{v}_j=v_{j+1}$ and $e{v}_j=j(m-j+1)v_{j-1}$. Its transpose equals $(-1{)}^{m}B$, so it is a symmetric bilinear form when $m$ is an even number, and an alternating bilinear form when $m$ is an odd integer. The Weyl complete reducibility theorem gives a nondegenerate invariant form on any finite-dimensional module by taking the orthogonal direct sum of these forms.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

Use the classification of finite-dimensional sl2 representations. In the irreducible representation $V(m)$, choose $v_j=f^j{v}_0$, $0\le j\le m$, sowith vectors beyond the endpoints interpreted as zero. The invariant form on an irreducible sl2 module isIts anti-diagonal entries are nonzero, so it is nondegenerate. Lie-invariant bilinear form invariance under $h$ follows from the sum of the two weights. For $f$, the two potentially nonzero terms are $(-1{)}^{j+1}$ and $(-1{)}^j$. For $e$, their coefficients coincide when $j+k=m+1$, and their signs are opposite. These checks prove invariance under the generators of the sl2 Lie algebra.

Interchanging $j,k$ multiplies the form by $(-1{)}^m$, givingEquivalently it is a symmetric bilinear form in odd integer dimension and an alternating bilinear form in even number dimension. The Weyl complete reducibility theorem expresses any finite-dimensional representation as a direct sum of these irreducibles. Give each summand the displayed form and make different summands orthogonal. The resulting form is invariant and nondegenerate. On a reducible representation with both parities, this orthogonal sum need not itself be symmetric or alternating; the dichotomy in the preceding part required irreducibility.

For the finite-dimensional Lie algebra representation specified in the PDF, a Lie-invariant bilinear form satisfiesThe dual Lie algebra representation has action $(x\varphi )(w)=-\varphi (xw)$, so the map $T_B:V\to V^{\ast }$ defined by $T_B(v)=B(v,-)$ is an intertwining operator. If $V$ is irreducible and $B\ne 0$, its kernel is zero. Since $V,V^{\ast }$ have equal finite dimension, $T_B$ is an isomorphism, and $B$ is a nondegenerate bilinear form.

For $B_1\ne 0$, the composition $T_{B_1}^{-1}{T}_{B_2}$ is a representation endomorphism. Over an algebraically closed field, the Schur lemma makes it scalar, provingTransposing $B$ gives another Lie-invariant bilinear form. For $B\ne 0$, write $B^T=\mu B$; transposing twice gives ${\mu }^2=1$. In field characteristic different from two, this means $\mu =1$ or $\mu =-1$. Thus the form is a symmetric bilinear form or an alternating bilinear form, respectively. The zero form has both properties. In the second case $B(v,v)=-B(v,v)$ forces $B(v,v)=0$, using the same characteristic assumption.

Choose the chain basis $v_0,\dots ,v_6$ from the preceding part, normalized by $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. The associated sl2 Lie algebra strings give $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$ and $e_2{v}_2=v_1$; other raising actions used below vanish because their proposed weights do not occur.

In the exterior square, $v_0\wedge v_1$ is a highest-weight vector of weight $2{\omega }_1-{\alpha }_1={\omega }_2$: $e_1$ produces $v_0\wedge v_0=0$, and $e_2$ kills both factors. Alsois a nonzero highest-weight vector of weight ${\omega }_1$. Its two $e_1$ terms cancel, and its $e_2$ image is zero. The Weyl complete reducibility theorem supplies irreducible summands with those weights; their dimensions $14+7=21$ exhaust $\dim {\Lambda }^{2}V$. ThusFor the symmetric square, $v_0\otimes v_0$ is a highest-weight vector of weight $2{\omega }_1$, giving a 27-dimensional summand. The longest Weyl-group element of $G_2$ acts as minus the identity, so the highest weight of a dual representation is $-w_0{\omega }_1={\omega }_1$: $V$ is self-dual. A nonzero intertwiner $V\to V^{\ast }$ gives a nondegenerate Lie-invariant bilinear form. It is symmetric, since an alternating bilinear form cannot be nondegenerate in odd dimension. Its inverse is a nonzero invariant vector in $S^{2}V$, supplying the scalar summand. Since $27+1=28=\dim S^{2}V$,These are the exterior and symmetric squares of the seven-dimensional G2 representation; all highest-weight and dimension claims are specified rather than relying on an unstated cross-product identity.

The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum: