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Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.

It is enough to assume that $h_k:{\pi }_k(X)\otimes \mathbb{Q}\to H_k(X;\mathbb{Q})$ is surjective for every $k\ge 2$; an integrally surjective Hurewicz homomorphism certainly has this property. First give the minimal-model argument suggested by the hint. Let $A=(\Lambda V,d)$ be a simply connected Sullivan minimal model, with $V$ in degrees at least two. Put $A^{+}={⨁}_{k>0}{A}^k$. Its indecomposable quotient of an augmented algebra isMinimality makes the induced differential on $V$ zero, so projection is a chain map $q:A^{+}\to (V,0)$. The Sullivan model identification says that the induced mapis dual to rational Hurewicz homomorphism. Under the usual finite-type duality convention, the assumed surjectivity makes $q_{\ast }$ injective.

A product of two positive cohomology classes has a decomposable cocycle representative, so its image under $q_{\ast }$ is zero. Injectivity therefore givesThis deduction uses the hypothesis; zero cup products by themselves would not yet prove formality of a topological space.

Choose a homogeneous cohomology basis $h_{\lambda }$ and cocycle representatives $z_{\lambda }\in A$. Their indecomposable parts $q(z_{\lambda })$ are linearly independent. Extend them to a graded basis of $V$ and replace the corresponding generators by the actual cocycles $z_{\lambda }$. This is a valid change of free algebra generators: each $z_{\lambda }$ differs from its indecomposable part by products of generators of strictly smaller degree, so degree induction constructs the inverse substitution. Call these closed generators $C$, and the remaining generators $N$.

Define a map of graded algebrasIts target has zero differential. Every differential in $A$ is decomposable, and all products of positive target classes vanish, so $\varphi$ is a chain map. If $z$ is any positive cocycle, its indecomposable part is in the span of the $q(z_{\lambda })$, because these form the image of $q_{\ast }$. Write $q(z)={\sum }_{\lambda }{c}_{\lambda }q(z_{\lambda })$. Then $\varphi (z)={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$, while injectivity of $q_{\ast }$ gives $[z]={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$. Thus $\varphi$ induces the identity on cohomology and is a quasi-isomorphism. This proves that $X$ is a formal space.

The rational wedge sum assertion has a direct topological proof, which also avoids finite-dimensional duality restrictions in the preceding model argument. Choose a homogeneous basis $b_{\lambda }$ of ${\tilde{H}}_{\ast }(X;\mathbb{Q})$. By surjectivity, each basis class is the image under the Hurewicz homomorphism of an element of ${\pi }_{n_{\lambda }}(X)\otimes \mathbb{Q}$. In $X_{\mathbb{Q}}$ this element is represented by an actual sphere map $S^{n_{\lambda }}\to X_{\mathbb{Q}}$. Simple connectivity and $H_1=0$ ensure $n_{\lambda }\ge 2$. The wedge sum of these maps givesand induces an isomorphism of rational homology by the choice of basis. The rational Whitehead theorem now makes it a rational homotopy equivalence, givingA wedge sum of spheres is a formal space. For each individual sphere, its augmented minimal model maps by a quasi-isomorphism to its sphere cochain algebra, and also to its cohomology algebra, by killing the extra odd generator in the even-sphere case. Take based simplicial models of the spheres. Their wedge sum is a simplicial pushout over the common basepoint, so the algebra of rational polynomial differential forms is the corresponding fibre product of augmented sphere algebras. Taking fibre products of these model maps gives a quasi-isomorphism zigzag to the square-zero cohomology algebra. This also works for infinitely many sphere summands: the positive parts are products of the individual complexes, and products of vector-space exact sequences over a field are exact. In the finite case its minimal resolution is the minimal model of a wedge of simply connected spheres. Thus the topological proof establishes both conclusions without imposing finite rank on the sphere basis. For the zero reduced-homology case the wedge sum is a point. Both the minimal-model construction and the explicit sphere map explain why the surjectivity hypothesis is decisive.

The homotopy fibre is the homotopy pullback of the path fibration over the target sphere along $f$. The Sullivan fibre-model theorem computes it by taking a relative acyclic model of that path fibration and tensoring it over the target model with the source model. Here the construction can be written out and checked directly.

Adjoin $a$ of degree $2n-1$ and $b$ of degree $4n-2$ to the sphere model, withThis is a differential, because $D(v-au)=u^2-u^2=0$. It is an acyclic relative Sullivan algebra: putting $v^{\prime }=v-au$ makes its two pairs $Da=u$ and $Db=v^{\prime }$. The first is a polynomial generator killed by an odd generator; the second is an odd generator killed by an even polynomial generator. Each pair has cohomology only $\mathbb{Q}$, the second by the ordinary polynomial antiderivative calculation in characteristic zero. Hence this models the contractible path space, and its augmentation models the chosen basepoint.

Tensoring with the projective-space model via the map in part (i) givesThe last differential is necessary: using only $Db=f^{\ast }v$ would give $D^{2}b=x^{2n}\ne 0$. Make the invertible generator change $z=y-xa$, of degree $2n+1$. Since $x$ is even,For $n\ge 2$, all nonzero differentials are decomposable and the generator ordering $x,a,z,b$ is Sullivan. Thus no further minimization is needed:For example at $n=2$ this is $(\Lambda (x_2,a_3,z_5,b_6),Da=x^2,Db=xz)$.

The indecomposable generators of a simply connected finite-type minimal model are dual to its rational homotopy groups. Therefore, for $n\ge 2$,These four degrees are distinct. The exact homotopy sequence provides an independent check: ${\mathbb{CP}}^n$ has rational homotopy only in degrees $2,2n+1$, while $S^{2n}$ has it in degrees $2n,4n-1$. For $n\ge 2$ these source and target degrees do not overlap, so the source classes survive in the fibre and the target classes contribute their one-degree-lower connecting classes, giving the same four groups.

At $n=1$, the displayed relative fibre algebra is not minimal: $Da=x$ and $Db=z$ are linear pairs. Cancelling both acyclic pairs leaves $\mathbb{Q}$. AccordinglyThis agrees with ${\mathbb{CP}}^1\cong S^2$ and the degree-one map being a homotopy equivalence. The case distinction is required before reading homotopy groups from the generators.

Work first over ${\mathbb{F}}_2$. The mod-two cohomology ring of real projective space is $H^{\ast }({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2[a]/(a^{n+1})$, with $|a|=1$. A map missing a point has zero mod-two degree of a map between closed manifolds, so $f^{\ast }(a^n)=0$. The group in degree one has only two elements, so $f^{\ast }a$ is either zero or $a$. The latter would imply $f^{\ast }(a^n)=(f^{\ast }a{)}^n=a^n\ne 0$, a contradiction. Therefore $f^{\ast }a=0$, and multiplicativity makes $f^{\ast }$ zero in every positive cohomological degree. Duality of homology and cohomology over a field consequently makes $f_{\ast }$ zero in every positive degree with mod-two coefficients.

For integral coefficients, cellular homology of real projective space gives $H_k\cong \mathbb{Z}/2$ in odd degrees $0<k<n$, zero in even degrees $0<k<n$, and a top group $\mathbb{Z}$ if $n$ is odd and zero if $n$ is even. For each torsion group, the natural reduction map to mod-two homology is injective by the universal coefficient theorem for homology; naturality and the preceding vanishing imply that its integral induced map is zero. When $n$ is odd, the integral top-degree map is also zero because the nonsurjective map has integer degree of a map between oriented manifolds zero. All groups above dimension $n$ vanish. Hence $f_{\ast }$ is zero on all positive integral homology groups as well.

For $n\ge 2$, ${\pi }_1({\mathbb{RP}}^n)=\mathbb{Z}/2$, and its abelianization is already $H_1({\mathbb{RP}}^n;\mathbb{Z})$. The vanishing just proved therefore implies $f_{\ast }{\pi }_1=0$. By the lifting criterion for a covering space, $f$ lifts through the double covering space $q:S^n\to {\mathbb{RP}}^n$ to a map $\tilde{f}:{\mathbb{RP}}^n\to S^n$ with $q\tilde{f}=f$.

Choose a point $y$ missed by $f$. Both points of $q^{-1}(y)$ are missed by $\tilde{f}$. In particular the lift has image in $S^n\setminus \{z\}\cong {\mathbb{R}}^n$, for either such point $z$. This punctured sphere is contractible, so contraction there gives a homotopy from $\tilde{f}$ to a constant map into $S^n$. Composing with $q$ contracts $f$. ThusThis proves the nonsurjective self-map of real projective space is null-homotopic conclusion, and also shows that all positive-degree induced maps vanish with any coefficients.

Let the integer degree of a map between oriented manifolds be $d\ne 0$, so $f_{\ast }[X]=d[Y]$. For any nonzero $\alpha \in H^k(Y;\mathbb{Q})$, nondegeneracy of the Poincare duality pairing supplies $\gamma \in H^{n-k}(Y;\mathbb{Q})$ with $⟨\alpha ⌣\gamma ,[Y]⟩\ne 0$. Naturality of the cup product and its evaluation givesThus $f^{\ast }\alpha \ne 0$, proving injectivity of pullback on rational cohomology. By the universal coefficient theorem for cohomology, these finite-dimensional rational cohomology groups are the duals of the corresponding homology groups, and the dual of $f_{\ast }$ is $f^{\ast }$. Injectivity of the dual map therefore provesOne can see the surjection constructively. Given $u\in H_k(Y;\mathbb{Q})$, choose its Poincare dual $\eta \in H^{n-k}(Y;\mathbb{Q})$ with $\eta \cap [Y]=u$. Then $v=d^{-1}(f^{\ast }\eta \cap [X])$ satisfies $f_{\ast }v=d^{-1}(\eta \cap f_{\ast }[X])=u$ by naturality of the cap product. Division by $d$ is the reason rational coefficients suffice.

Use the short exact sequence of coefficient groupsThe groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map $r_{\ast }$. Define the Bockstein homomorphism to be its connecting homomorphism.

Explicitly, represent $x$ by a mod-$p$ cocycle $a$ and lift it to an integral singular cochain $A$. Since $a$ is a cocycle, $\delta A$ is divisible by $p$. ThenThis is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod-$p^2$ coboundary of $A$ lies in the image of $j$. The quotient cochain is a cocycle, since ${\delta }^{2}A=0$ and integral cochains have no $p$-torsion. Replacing $A$ by another lift adds $pC$ and changes the quotient by $\delta C$ modulo $p$; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.

A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on $H^0(X;{\mathbb{F}}_p)$ for every space.

For a nonzero example in every positive degree $i$, take the Moore space $X=S^i{\cup }_p{e}^{i+1}$, attaching the cell by a map of mapping degree $p$. Its positive-degree integral cellular cochain complex has the differential $\mathbb{Z}\stackrel{p}{\to }\mathbb{Z}$ from degree $i$ to $i+1$; when $i=1$, the preceding differential from degree zero is zero. Modulo $p$ the displayed differential vanishes, and $H^i(X;{\mathbb{F}}_p)=H^{i+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Lift the degree-$i$ generator to the cellular cochain with value one. Its coboundary has value $p$, so dividing by $p$ gives the generator in degree $i+1$. Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore $\beta$ is nonzero, indeed an isomorphism, in the requested degree for every prime $p$.

Finally take integral lifts $A,B$ of cocycles representing $x\in H^i(X;{\mathbb{F}}_p)$ and $y\in H^j(X;{\mathbb{F}}_p)$. Write $\delta A=p{A}_1$ and $\delta B=p{B}_1$. The cochain cup product obeys the graded Leibniz identity, soDivide by $p$, reduce modulo $p$ and pass to cohomology. This proves the Bockstein derivation rule

For a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$, putThese arbitrary families of germs form a sheaf, and restrictions are projections. Extending a family by zero proves that $C(\mathcal{F})$ is flasque. Sending a section to all its germs gives an injective sheaf map $\mathcal{F}\to C(\mathcal{F})$, since a section with every germ zero vanishes locally and hence globally.

Set $Q^{-1}=\mathcal{F}$, $I^j=C(Q^{j-1})$ and $Q^j=\mathrm{coker}(Q^{j-1}\to I^j)$, using the sheaf cokernel. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the Godement resolutionIt is exact by construction, with flasque terms. Define sheaf cohomology by the cohomology of its global-section complex:The resolution principle for sheaf cohomology identifies this with the computation using any flasque resolution.

The elementary input for acyclicity is the flasque-kernel section-lifting lemma: in a short exact sequence with flasque kernel, sections of the quotient lift over every open subset. Consequently a quotient of two flasque sheaves is flasque, because a quotient section lifts and that lift extends. If $\mathcal{F}$ itself is flasque, induction through $0\to Q^{j-1}\to I^j\to Q^j\to 0$ makes every $Q^j$ flasque. The same lifting lemma makes these sequences exact on global sections. Hence the global-section complex has no positive-degree cohomology, provingThis deduction uses the lifting lemma, not the desired cohomological vanishing as an assumption.

For the rational-section construction, add two representative sections after restricting to the intersection of their dense domains. Multiply by a representative rational function in the same way. Finite intersections of dense open sets are dense, and further restricting both representatives does not change their resulting class. The module axioms hold on their common domain. Thus rational sections form a module over $\mathrm{Rat}(X)$; irreducibility is not needed for this module statement. When $X$ is irreducible, $\mathrm{Rat}(X)$ is the function field $K=k(X)$.

Suppose now that $\mathcal{F}$ is locally free and $X$ irreducible. A germ at $P$ is represented on a nonempty open neighbourhood, which is dense, so it defines a rational section. Two representatives of the germ agree on a neighbourhood of $P$ and therefore give the same class. If the class is zero, trivialize $\mathcal{F}$ on a neighbourhood of $P$. Its coefficient functions vanish on a dense open subset; regular functions on a reduced irreducible variety with this property vanish identically. The germ is therefore zero. This proves the stalk inclusion into rational sections of a locally free sheaf:Local freeness matters: a nonzero torsion germ can vanish on a dense open set.

On an irreducible algebraic curve, every proper closed subset is finite, and every open subset is quasi-compact because the Zariski space is Noetherian. Let $R=\mathrm{Rat}(\mathcal{F})$. The constant sheaf $\mathcal{R}(\mathcal{F})$ has sections $R$ on every nonempty open set: such opens are irreducible and hence connected. DefineRestriction drops the components outside the smaller open set. For a compatible family over an open cover of $U$, take a finite subcover by quasi-compactness. The union of the finite supports of these finitely many sections is finite. Their agreeing components therefore glue to a unique finite-support tuple on $U$. This verifies the sheaf gluing axiom, including arbitrary covers, rather than merely asserting that a presheaf direct sum is always a sheaf.

A rational section is regular on some dense open set, whose complement on a curve is finite. Its classes $[s{]}_P\in R/{\mathcal{F}}_P$ therefore have finite support, defining a sheaf map $\mathcal{R}(\mathcal{F})\to \mathcal{P}(\mathcal{F})$. At $P$, the first sheaf has stalk $R$. The second has stalk $R/{\mathcal{F}}_P$: any finite tuple can be restricted to a neighbourhood omitting all its other support points, and its $P$-component is unaffected by such restriction. The resulting sequence on stalks isExactness of sheaves can be tested on stalks, so this proves the rational principal-parts resolution on an algebraic curveBoth right-hand sheaves are flasque: restrictions for the constant rational-section sheaf are identities between nonempty opens, and restrictions for the rational principal-parts sheaf are projections, with extension by zero. ConsequentlySurjectivity on stalks has not been confused with surjectivity on global sections.

For an explicit failure of global surjectivity take $X={\mathbb{P}}^1$ and $\mathcal{F}=\mathcal{O}(-2)$. Use coordinate $z=X_1/X_0$ and frame $e_0=X_0^{-2}$ on the finite chart. At infinity use $w=1/z$ and $e_{\infty }=X_1^{-2}=z^{-2}{e}_0$. Prescribe the rational principal part of $z^{-1}{e}_0$ at zero and zero at every other point. If a rational section $r(z)e_0$ realized this tuple, then $r-z^{-1}$ would be regular at zero and $r$ would be regular at every other finite point. Thus $r=z^{-1}+p(z)$ with $p\in k[z]$. Regularity at infinity would requireto be regular at $w=0$. Its $w^{-1}$ term cannot cancel with any term of the polynomial contribution, whose powers are at most $-2$. This is impossible. Hence the global rational-principal-part map need not be surjective.

For an affine curve it is always surjective, and here is a direct construction. Write $X=\mathrm{Spec}A$, $K=\mathrm{Frac}A$ and $\mathcal{F}=\tilde{M}$. By the affine module sheaf construction, $R=M{\otimes }_{A}K=:M_K$ and ${\mathcal{F}}_P=M_{{\mathfrak{m}}_P}$. Choose representatives $v_P\in M_K$ for a prescribed finite tuple, and choose one nonzero $a\in A$ with $a{v}_P\in M$ for all of them. The zero set $T=V(a)$ is finite. The ring $A/(a)$ is Artinian, since it is Noetherian of dimension zero, and its Artinian decomposition into local factors givesPrescribe the class of $a{v}_P$ in the indicated component for each requested point in $T$, and zero in the other components. A requested point outside $T$ already has zero principal part because $a$ is a unit there. Lift the tuple to $w\in M$. Then $s=w/a\in M_K$ satisfies $s-v_P\in M_{{\mathfrak{m}}_P}$ at every requested point in $T$, and is regular at every other point of $T$. Outside $T$ it is regular because $a$ is invertible. It therefore realizes exactly the prescribed tuple. This proves affine principal-parts interpolation for a locally free sheaf and givesNo nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.

In the cohomology basis $h_1,\dots ,h_n$ supplied by the projective lines in the summands, the intersection form isFor each permutation $\sigma \in S_n$, the linear map $h_i\mapsto h_{\sigma (i)}$ preserves this pairing. Distinct permutations induce distinct maps, and their compositions agree with permutation composition. Hence the permutation matrices give a subgroup $G\cong S_n$ of the isometry group of $Q_X$.

To realize these isometries smoothly, describe $X$ as a punctured $S^4$ with $n$ punctured copies of ${\mathbb{CP}}^2$ attached to its boundary spheres. An orientation-preserving ambient isotopy of $S^4$ can interchange any two chosen disjoint attachment balls while carrying their parametrizations along with them: move their centers along disjoint arcs and extend the motion to small balls. There is room to make these arcs disjoint in dimension four. Extend the endpoint diffeomorphism across the two exchanged punctured copies using the chosen identifications of those identical oriented summands; use the identity on the remaining copies. Collar coordinates make the maps fit smoothly along the gluing spheres. This exchanges the two line classes without changing their signs. Since transpositions generate $S_n$, composing these diffeomorphisms realizes each element of $G$. Pullback permutes the cohomology basis by the inverse permutation, which still realizes the same subgroup and all its elements. These are permutation diffeomorphisms of identical connected summands.

For the adjoint real bundle $\mathrm{ad}E$, the ASD deformation complex iswhere $d_A^{+}=P_{+}{d}_A$ and $P_{+}=(1+\ast )/2$. On an adjoint section $\xi$, the vector-bundle curvature identity gives $d_A^2\xi =[F_A,\xi ]$. Since $P_{+}$ acts only on the two-form part and $F_A^{+}=0$,This proves it is a complex. Its three cohomology spaces are the infinitesimal stabilizers $H_A^0$, infinitesimal deformations modulo gauge $H_A^1$, and obstructions $H_A^2$.

The associated gauge-fixed elliptic differential operator isThe ASD deformation index formula isHere the Atiyah-Singer index theorem is used for the formula, with $p_1(\mathrm{ad}E)=-4c_2(E)$ and $e(E)=c_2(E)[X]$. For the simply connected negative-definite base in this question, $b_1=b^{+}=0$, so $\mathrm{ind}{D}_A=8e(E)-3$. The alternating Euler characteristic of the three-term complex is the negative of this operator index; stating which index is being used avoids a sign ambiguity.

The three identifications areThey match the same lower-dimensional theory, including its moduli of a string compactification, charged states and extended branes. Different weakly coupled descriptions occupy different limits of its moduli space.

For the M-theory circle duality, write ${\ell }_s=\sqrt{{\alpha }^{\prime }}$ and let ${\ell }_p$ be the eleven-dimensional Planck length. The standard length dictionary isThe circle radius is the type IIA superstring theory coupling modulus. The eleven-dimensional metric supplies the ten-dimensional metric, dilaton and Ramond–Ramond potential $C_1$ through Kaluza-Klein theory; the three-form supplies $C_3$ and the Neveu–Schwarz two-form $B_2$ by $A_3=C_3+B_2\wedge dy$. Thus the field content also agrees with type IIA supergravity.

The M2-brane and M5-brane tensions are $T_2=(2\pi {)}^{-2}{\ell }_p^{-3}$ and $T_5=(2\pi {)}^{-5}{\ell }_p^{-6}$. Wrapping one spatial direction multiplies the effective brane tension by $2\pi R_{11}$. Consequently:

These tension and charge relations explain why increasing $g_A$ reveals an eleventh dimension. The simple circle dictionary describes massless type IIA supergravity; a D8-brane involves its massive extension and Romans mass.

For the M-theory torus duality, the complex structure of the torus is the IIB axion-dilatonIts area supplies the remaining radius modulus. In a common nine-dimensional mass normalization, an M2-brane wrapped on the entire torus has mass $T_2\mathcal{A}$, identified with $1/R_B$; hence $R_B$ is inversely related to the torus area in that normalization. Radii quoted in different string or Einstein metrics also include their Weyl rescaling. A rectangular torus with cycle radii $R_{10},R_{11}$ has $g_B=R_{11}/R_{10}$. Equivalently, first reduce to IIA and use T-duality:where $R_A,R_B$ in this formula use the corresponding string metrics. Shrinking the torus area at fixed $\tau$ decompactifies the IIB circle. The torus mapping class group $SL(2,\mathbb{Z})$ becomes IIB S-duality.

An M2-brane wrapped on a primitive $(p,q)$ one-cycle gives a (p,q) string, whose tension is the membrane tension times the cycle length before conversion to the IIB metric. Momentum on the torus matches winding of these strings on the IIB circle. An M5-brane wrapped on the whole torus gives an unwrapped D3-brane; an unwrapped M2-brane gives a D3-brane wrapped on the IIB circle. An M5-brane wrapped on one torus cycle gives a (p,q) five-brane wrapped on the IIB circle. The unwrapped M5-brane is the magnetic dual of the IIB circle's Kaluza-Klein mode, represented by its Kaluza-Klein monopole five-brane. These identifications match effective spatial dimensions as well as electric and magnetic charges.

For the M-theory K3 duality, the common noncompact spacetime has seven dimensions and sixteen supersymmetry generators. A Ricci flat K3 surface has a positive three-plane in its second real cohomology, equipped with the K3 intersection lattice of signature $(3,19)$. The unit-volume moduli space of Ricci-flat K3 metrics has local formwith a discrete lattice identification and one additional positive volume modulus. This is the heterotic Narain moduli space on $T^3$, together with the seven-dimensional coupling: its torus metric has six parameters, its antisymmetric tensor three, and its sixteen gauge Wilson lines forty-eight, giving $6+3+48=57$ before the dilaton.

The coupling–volume relation can be checked dimensionally rather than guessed. A wrapped M5-brane gives a heterotic fundamental string with tension $T_{5}V$, so its squared string length is proportional to ${\ell }_p^6/V$. The seven-dimensional gravitational coupling is proportional to ${\ell }_p^9/V$. Their dimensionless ratio givesThus the large-volume M-description corresponds to strong heterotic coupling. Overall constants depend on the normalization of the lower-dimensional coupling.

There are twenty-two gauge fields from reducing $A_3$ on the twenty-two harmonic two-forms of K3, matching the heterotic sixteen gauge vectors plus three metric and three two-form vectors. The remaining seven-dimensional three-form is dual to the heterotic two-form. A wrapped M2-brane gives an electrically charged particle in the shared charge lattice, matching heterotic momentum, winding and gauge charges; an M5-brane on a two-cycle gives its magnetic three-brane. An M5-brane wrapping all of K3 gives the heterotic string, while an unwrapped M2-brane corresponds to a heterotic NS5-brane wrapped on $T^3$. At suitable collapsing two-cycles, charged membrane states become massless and yield ADE gauge enhancement. The geometry accounts for both generic abelian charges and nonabelian enhancement. There is no extra K3 two-form modulus in the M-description: its fundamental potential is a three-form, and $b_3(K3)=0$.