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The Borsuk-Ulam theorem says that recording $n$ continuous real-valued measurements on an $n$-dimensional unit sphere must give identical records somewhere on opposite sides. The measurements can vary wildly; continuity alone forces a coincidence at an antipodal pair.

Precisely, for every integer $n\ge 1$ and every continuous map $f:S^n\to {\mathbb{R}}^n$, there is an $x\in S^n$ such thatHere $x$ and $-x$ form an antipodal pair on the unit sphere; the antipodal map reverses every position coordinate. The target is the Euclidean space ${\mathbb{R}}^n$. Both points have the same entire output, so every one of the $n$ measurements agrees at this common pair. See Hatcher, Algebraic Topology, Corollary 2B.7, p. 176.

Circle example. Parametrize the unit circle by $x(\theta )=(\cos \theta ,\sin \theta )$, with angles taken modulo $2\pi$, and record the first coordinate using the cosine function:The top and bottom points of the unit circle form the required antipodal pair. Their first coordinates agree, while their second coordinates are $1$ and $-1$.

Sphere example. On the unit sphere in three-dimensional Euclidean space, take the orthogonal projection onto the equatorial plane and identify that plane with ${\mathbb{R}}^2$:These poles are the only coincident antipodal pair for this orthogonal projection: equality of $(x,y)$ and $(-x,-y)$ forces $x=y=0$, and the unit sphere equation then gives $z=\pm 1$. The Borsuk-Ulam theorem guarantees an antipodal pair for every continuous map into ${\mathbb{R}}^2$, even when no convenient projection formula is available.

Equal output need not mean equal position coordinates. Keeping all three coordinates gives a continuous map $i:S^2\to {\mathbb{R}}^3$ defined by $i(x,y,z)=(x,y,z)$, the restriction of the identity map of ${\mathbb{R}}^3$. Its outputs at an antipodal pair are always distinct, since $u=-u$ would force $u=0$, which is outside the unit sphere. Thus replacing the target ${\mathbb{R}}^2$ by ${\mathbb{R}}^3$ destroys this guarantee.

Why the circle case follows from the intermediate value theorem. For an arbitrary continuous map $f:S^1\to \mathbb{R}$, write $F(\theta )=f(x(\theta ))$ using the unit circle parametrization above. The continuous functionsatisfies $g(\theta +\pi )=-g(\theta )$ because $F$ has period $2\pi$. If $g(0)=0$, the first antipodal pair already works. Otherwise $g(0)$ and $g(\pi )$ have opposite signs, and the intermediate value theorem supplies a zero between them. At that zero, the two antipodal values of $f$ agree. In the cosine function example, $g(\theta )=2\cos \theta$. This short argument handles only $n=1$. See Hatcher, Algebraic Topology, discussion of Theorem 1.10, p. 32.

An ideal weather interpretation. Model the surface of Earth as a sphere. At one fixed instant, suppose temperature $T$ and pressure $P$ are globally defined continuous scalar measurements on that sphere. Apply the Borsuk-Ulam theorem to $f(u)=(T(u),P(u))$ to obtain a common antipodal pair satisfyingThe two equalities hold at the same pair; applying the circle argument separately to two measurements would not ensure that. This is a conclusion about the ideal continuous model, with both measurements taken at the same instant, rather than an assertion that an exactly matching pair has been empirically measured. Matoušek gives this motivating interpretation and explicitly cautions that physical temperature need not be continuous.^[ref]

Equivalent formulations. A continuous map $h:S^n\to {\mathbb{R}}^n$ is odd when $h(-x)=-h(x)$, extending the symmetry of an odd function. The Borsuk-Ulam theorem equivalently says every such map has a zero. For a general $f$, the difference $h(x)=f(x)-f(-x)$ has this symmetry; conversely, equal antipodal outputs of an odd map must be zero. Another equivalent formulation is that no odd map between spheres $S^n\to S^{n-1}$ exists. Normalizing a nonvanishing odd map by its Euclidean norm would produce such an odd map between spheres. These formulations are useful starting points for algebraic topology; the general proof is beyond the circle argument here.^[ref]

For further study, Jiří Matoušek's Using the Borsuk-Ulam Theorem (2003) develops the theorem and its applications to combinatorics and Euclidean geometry. An early publication is Karol Borsuk's 1933 paper Drei Sätze über die n-dimensionale euklidische Sphäre.

A cover of $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The equivalent version for open sets follows by shrinking finite open covers on a compact metric space; the reverse implication follows by enlarging antipodal-pair-free closed sets slightly. This is a covering formulation of the Borsuk-Ulam theorem.

The principal symbol of an odd-order scalar differential operator is an odd map from the frequency sphere to the complex plane. The Borsuk-Ulam theorem forces a zero on a two-dimensional sphere, contradicting ellipticity. Restriction to three variables proves the same obstruction in every greater dimension.

For a finite open cover $U_0,\dots ,U_d$ of $S^d$, take continuous nonnegative functions ${\varphi }_i$ summing to one and positive only in $U_i$. Such a partition of unity comes from normalizing the distance to a set functions $d(x,S^d\setminus U_i)$. The Borsuk-Ulam theorem applied to $({\varphi }_1,\dots ,{\varphi }_d)$ gives equal values at an antipodal pair. The sum condition makes ${\varphi }_0$ equal too, and some common positive value places both points in one $U_i$.

The standard simplicial decomposition of a sphere $F^n$ is the boundary of the cross-polytope with vertices $\pm e_1,\dots ,\pm e_{n+1}$, identified radially with $S^n$. A face chooses distinct coordinate indices and one sign for each; it never contains an antipodal pair.

A regular antipodal triangulation of a sphere is invariant under the antipodal map and has the nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ as subcomplexes. In particular each equator separates its sphere into two triangulated hemispheres. These conditions are preserved by barycentric subdivision.

For a simplicial map $f:F\to F^n$, a positive alternating simplex of dimension $j$ has distinct image labelsThe opposite alternating pattern is negative; all other simplices, including collapsed images, are neutral.

To prove the antipodal alternating-simplex parity lemma, let $p_j$ count positive $j$-simplices in the equatorial $S^j$. There is exactly one positive vertex in $S^0$, so $p_0=1$. A $j$-simplex has an odd number of positive alternating facets precisely when it is positive or negative alternating. Indeed an alternating sequence has exactly one positive alternating deletion; a nonalternating sequence has either zero or two. For a collapsed image, a positive facet can occur only when exactly one label is repeated; deleting either copy gives two such facets.

Count incidences of positive $(j-1)$-faces with $j$-simplices in one closed hemisphere. Interior faces contribute twice and equatorial faces once. The count modulo two is therefore $p_{j-1}$. It is also the number of positive or negative $j$-simplices in that hemisphere modulo two. Because $f$ is antipodal, each alternating pair $\sigma ,-\sigma$ has one positive and one negative member, with one member in each hemisphere. ThusInduction gives odd $p_k$, proving the existence of a positive $k$-simplex. In particular, an antipodal simplicial map $F\to F^{k-1}$ is impossible: a positive $k$-simplex would require $k+1$ distinct coordinate labels but only $k$ are available.

The Borsuk-Ulam theorem states that every continuous map $g:S^k\to {\mathbb{R}}^k$ has some $x$ with $g(x)=g(-x)$. First rule out a continuous antipodal map $h:S^k\to S^{k-1}$ for $k\ge 1$. By uniform continuity, take a sufficiently fine regular barycentric subdivision so that endpoints of each edge have images less than $2/\sqrt{k}$ apart. Label a vertex by $\mathrm{sgn}(h_i)e_i$, where $i$ is the smallest index maximizing $|h_i|$. Since $h$ is unit length, this maximum is at least $1/\sqrt{k}$; two adjacent vertices cannot have opposite labels. The labels therefore define an antipodal simplicial map to $F^{k-1}$, contradicting the preceding result.

If the stated Borsuk-Ulam theorem failed, the mapwould be just such a continuous antipodal map. This proves the required version; $k=0$ is immediate.

For the Kneser conjecture, equivalently Lovász theorem on Kneser graphs, the claim isThe Kneser graph joins disjoint $r$-subsets of $[N]$. Let $d=N-2r+1$. Colour each $r$-set by its least element in $[d]$ when it has one, and otherwise by one final colour. A colour class indexed by $i$ is an intersecting family because all its members contain $i$; the final class lies in a $(2r-1)$-set and is also intersecting. This proves the upper bound $d+1$.

For the lower bound, first deduce the open-cover Lusternik-Schnirelmann-Borsuk theorem. An open cover $U_0,\dots ,U_d$ of $S^d$ admits a continuous partition of unity ${\varphi }_0,\dots ,{\varphi }_d$ subordinate to it. Apply the Borsuk-Ulam theorem to $({\varphi }_1,\dots ,{\varphi }_d)$. Equal values at $x,-x$ and ${\sum }_i{\varphi }_i=1$ make all values equal. At least one common value is positive, so some $U_i$ contains an antipodal pair.

Now suppose the $r$-sets could be coloured with only $d$ colours, each an intersecting family. Choose $N$ points $v_1,\dots ,v_N\in S^d$ such that every $d+1$ of their vectors are linearly independent. For example, normalize $(1,t_i,\dots ,t_i^d)$ for distinct real $t_i$; the Vandermonde determinant proves this property. The general-position hemisphere count says that at most $d$ points lie on any equator. Since $N-d=2r-1$, at least one of the opposite open hemispheres determined by $x,-x$ contains an $r$-set of the labelled points.

Let $U_i$ be the open set of directions whose positive open hemisphere contains an $r$-set of colour $i$. It contains no antipodal pair, as such a pair would give two disjoint sets of that colour. The compact set $K=S^d\setminus {\bigcup }_{i=1}^d{U}_i$ also contains no antipodal pair: otherwise each of the two opposite hemispheres would have at most $r-1$ points, contradicting the count $2r-1$.

If $K$ is nonempty, compactness gives a positive set distance between $K$ and $-K$; an open neighbourhood $U_0$ of radius less than half that set distance still has no antipodal pair. If $K$ is empty, take $U_0=\varnothing$. In either case $U_0,U_1,\dots ,U_d$ is an open cover with no member containing an antipodal pair, contradicting the Lusternik-Schnirelmann-Borsuk theorem. Hence at least $d+1$ colours are required.

For integral singular cochains $a\in C^p(X)$ and $b\in C^q(X)$, define the cup product on an ordered simplex byThe same construction works over any commutative coefficient ring. Expanding the coboundary as the alternating sum of faces gives the Leibniz identityIndeed faces deleting vertices before the split occur in the first sum, faces after the split occur in the second, and the two terms at the common vertex cancel with opposite signs. Thus the product of cocycles is a cocycle. Replacing $a$ by $a+\delta c$ changes the product with a cocycle $b$ by $\delta (c⌣b)$; replacing $b$ by $b+\delta d$ changes the product with a cocycle $a$ by $(-1{)}^p\delta (a⌣d)$. Hence the product is independent of representatives and defines a well-defined product on cohomology classes.

With ${\mathbb{F}}_2={\mathbb{Z}}_2$ coefficients, the real-projective-space cohomology ring isFor $n>m\ge 1$, the pullback of the degree-one generator is either zero or $t$. The second choice would send the zero class $t^{m+1}$ to the nonzero class $t^{m+1}\in H^{m+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)$, contradicting the naturality of the cup product. Thus the generator pulls back to zero, and so do all its positive powers. For $m=0$ the target's reduced cohomology is already zero. Therefore every such map induces zero on reduced mod-two cohomology.

Suppose a continuous $f:S^n\to {\mathbb{R}}^n$ had $f(x)\ne f(-x)$ everywhere. Thenis a continuous odd map to $S^{n-1}$ and descends to $\bar{h}:{\mathbb{RP}}^n\to {\mathbb{RP}}^{n-1}$. For $n\ge 2$, a path from $x$ to $-x$ projects to a loop detected by the degree-one generator of the double cover. Its image lifts to a path from $h(x)$ to $-h(x)$, so is detected by the target's double cover as well. Thus ${\bar{h}}^{\ast }$ is nonzero in degree one, contradicting the preceding calculation. For $n=1$, an odd continuous map from the connected circle to $S^0$ is already impossible. The case $n=0$ is immediate. HenceThis proves the Borsuk-Ulam theorem rather than assuming it.

For a closed cover $A_1,\dots ,A_{n+1}$, assume no member contains an antipodal pair. Apply the just-proved theorem to $F(x)=(d(x,A_1),\dots ,d(x,A_n))$, using any metric on the sphere. Empty sets can be assigned a constant positive coordinate. It gives equal distances at $x,-x$. If either point belongs to an $A_i$ with $i\le n$, zero distance and closedness place both points in it, a contradiction. Hence neither lies in any of the first $n$ sets, and both must lie in $A_{n+1}$, another contradiction. This proves the Lusternik-Schnirelmann-Borsuk theorem for closed sets.

Four closed sets on $S^2$ need not contain an antipodal pair. Take the four unit vertices $v_i$ of a regular tetrahedron centered at zero and define $C_i=\{x:x\cdot v_i\ge x\cdot v_j\text{ for every }j\}$. These closed sets cover the sphere by maximizing the four scalar products. If $x$ and $-x$ belonged to $C_i$, both inequalities would force all four scalar products to be equal. Since $\sum v_i=0$, they would all be zero, and because the vertices span ${\mathbb{R}}^3$, $x$ would be zero. This is impossible on the sphere. Thus these four sets give the antipodal-free closed cover from simplex Voronoi cells counterexample.

Suppose an odd continuous map $f:S^2\to S^1$ existed. Restrict to the equator, parametrized by $e^{i\theta }$. Choose a continuous real lift $F$ with $f(e^{i\theta })=e^{iF(\theta )}$. Oddness gives $F(\theta +\pi )-F(\theta )=\pi +2\pi k$ for a fixed integer $k$, since the integer-valued difference is continuous. Thus $F(\theta +2\pi )-F(\theta )=2\pi (1+2k)$, and the equatorial map has odd degree of a continuous mapping, hence nonzero degree. But it extends over a hemisphere, which is a disk, so the equatorial loop is null-homotopic and has degree zero. This contradiction proves the dimension-two Borsuk-Ulam theorem in the stated form.

If a continuous injection $h:S^2\to {\mathbb{R}}^2$ existed, then $h(x)-h(-x)$ would never vanish. Normalization would definea continuous odd map, already ruled out. In particular the sphere cannot be homeomorphic to any planar subset.

The Lusternik-Schnirelmann-Borsuk theorem has the following two equivalent covering formulations. For every integer $d\ge 0$, a cover of the sphere $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The same assertion holds with open sets in place of closed sets. Thus $d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$. The dimension-zero case simply says that one set covering the two-point sphere contains both points.

Here is why the two versions agree. A finite open cover of a compact metric space admits a closed shrinking that still covers: sufficiently small closed balls subordinate to the open cover can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty closed subset $C$ of $S^d$ avoids antipodal pairs, compactness gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids antipodal pairs. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.

Another common equivalent formulation is the Borsuk-Ulam theorem: every continuous $f:S^d\to {\mathbb{R}}^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every continuous odd function $g:S^d\to {\mathbb{R}}^d$ has a zero. For example, if $d+1$ antipodal-pair-free open sets covered $S^d$, a subordinate partition of unity $({\varphi }_1,\dots ,{\varphi }_{d+1})$ would give the odd functionIt cannot vanish, since some ${\varphi }_i(x)>0$, whereas antipodal-pair-freeness forces ${\varphi }_i(-x)=0$. In the other direction, if an odd function $g:S^d\to {\mathbb{R}}^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a regular simplex centred at the origin in ${\mathbb{R}}^d$. For $d\ge 1$, the open sets $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an antipodal pair, contradicting the covering theorem.

Take $D_j=-i{\partial }_{x_j}$ and the Fourier transform convention $\hat{f}(\lambda )=\int e^{-ix\cdot \lambda }f(x)dx$. Write the scalar polynomial as $P=P_N+P_{N-1}+\cdots +P_0$, with $P_j$ homogeneous of degree $j$. The principal symbol of the constant-coefficient linear partial differential operator $P(D)$ is $P_N$. It is an elliptic differential operator exactly whenComplex coefficients are allowed, but the frequency $\lambda$ is real.

By compactness of the unit sphere, $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Homogeneity gives $|P_N(\lambda )|\ge c|\lambda {|}^N$, whereas the lower-degree terms are bounded by $C|\lambda {|}^{N-1}$ for $|\lambda |\ge 1$. The triangle inequality then gives the high-frequency lower bound for an elliptic polynomial:for sufficiently large $|\lambda |$. ThusThe Japanese bracket is $⟨\lambda ⟩=(1+|\lambda {|}^2{)}^{1/2}$. If $N=0$, a nonzero constant $P$ satisfies the same assertion directly.

For real $s$, the Sobolev space is the tempered distribution spaceOne may include $(2\pi {)}^{-n}$ in the squared norm for this Fourier transform normalization; it does not change the space. The Local Sobolev space iswhere multiplication of a distribution by a smooth function is followed by extension by zero.

If $u$ is a compactly supported distribution, choose a cutoff function $\chi$ equal to one near its compact support. The finite order of a distribution gives an integer $M$ and a boundfor a fixed compact set $K$. This also makes $u$ a tempered distribution. Its Fourier transform of a compactly supported distribution is the smooth functionThe Leibniz rule gives the last estimate, and parameter differentiation under the finite-order pairing proves smoothness of this Fourier transform. Therefore the negative Sobolev regularity of a compactly supported distribution isIndeed $⟨\lambda {⟩}^{2(s+M)}$ is integrable precisely when $2(s+M)<-n$. This proves the claimed existence of a sufficiently negative Sobolev space index.

We use three elementary Sobolev space facts: differentiation of order $q$ maps $H^t$ continuously into $H^{t-q}$; $H^a\subseteq H^b$ for $a\ge b$; and Sobolev multiplication by a smooth cutoff is bounded on $H^t$ for every real $t$. The first two follow directly from the frequency weights. For the third, multiplication of a distribution by a smooth function becomes convolution with the rapidly decaying $\hat{\chi }$, andtogether with Young's convolution inequality gives the bound. These facts apply after localization as well.

The high-frequency lower bound for an elliptic polynomial gives a useful global implication. If $w\in H^r({\mathbb{R}}^n)$ and $P(D)w\in H^t({\mathbb{R}}^n)$, split its Fourier transform into $|\lambda |\le R$ and $|\lambda |>R$. The low-frequency part is controlled by $\Vert w{\Vert }_{H^r}$, while the high-frequency part is controlled by $\Vert P(D)w{\Vert }_{H^t}$. Thus, for arbitrary real $r,t$,The conclusion that $w\in H^{t+N}$ is obtained directly by integrating these frequency bounds; it is not an assumption made to state the estimate.

For the cutoff bootstrap for local elliptic regularity, fix $U⋐X$. A cutoff function equal to one near $\overline{U}$ makes $u$ a compactly supported distribution after multiplication, so the preceding negative-index argument gives $u\in H_{\mathrm{loc}}^r(U)$ for some finite $r$. Suppose inductively that $u\in H_{\mathrm{loc}}^q(U)$. For any test function $\chi \in C_c^{\infty }(U)$,The Leibniz rule shows that this commutator has order at most $N-1$, with smooth functions as coefficients, all with compact support:A second cutoff function equal to one near $\mathrm{supp}\chi$ lets us apply the stated Sobolev space bounds to every term. Hence $[P(D),\chi ]u\in H^{q-N+1}({\mathbb{R}}^n)$, while $\chi P(D)u\in H^s({\mathbb{R}}^n)$. For $N\ge 1$, the global estimate yieldsThis holds for every such $\chi$, so it improves the Local Sobolev space index by one until $s+N$ is reached. A finite number of iterations starting at $r$ provesSince $U⋐X$ was arbitrary, the conclusion holds throughout $X$. For $N=0$, division by the nonzero constant $P$ proves it immediately. This is the asserted elliptic regularity, proved without assuming an initial nonnegative Sobolev space index.

A first-order example on ${\mathbb{R}}^2$ is the first-order Cauchy-Riemann operatorIt is an elliptic differential operator with complex coefficients. In one real variable, $D=-id/dx$ is already a first-order elliptic differential operator.

There is no scalar elliptic differential operator of odd order in three variables. If its order $N$ were odd, its principal symbol would satisfy $P_N(-\omega )=-P_N(\omega )$ on $S^2$. Regard the continuous mapThe Borsuk-Ulam theorem gives $f(\omega )=f(-\omega )$ for some $\omega$. Oddness then gives $f(\omega )=0$, contradicting ellipticity. This is the odd-order obstruction for scalar elliptic operators in at least three dimensions; restricting to a three-dimensional subspace proves the higher-dimensional case too. For real coefficients alone, the same obstruction follows from the intermediate value theorem along a path between antipodal points.

The scalar qualification matters. An elliptic system of differential equations can be first order in three variables: with the Pauli matrices, the matrix symbol $A(\lambda )={\sum }_{j=1}^3{\sigma }_j{\lambda }_j$ satisfies $A(\lambda {)}^2=|\lambda {|}^{2}1$ and is invertible for $\lambda \ne 0$. This does not contradict the scalar polynomial obstruction.

Assume (i). An antipodal map $g:S^n\to S^{n-1}$, followed by the inclusion $S^{n-1}\subset {\mathbb{R}}^n$, would be an antipodal map with no zero. Hence (i) implies (ii). Conversely, if an antipodal $f:S^n\to {\mathbb{R}}^n$ had no zero, thenwould be an antipodal map to $S^{n-1}$. Thus (ii) implies (i).

If $F:B^n\to S^{n-1}$ is antipodal on the boundary, regard $S^n$ as two copies of $B^n$ glued along their boundary. Use $F(x)$ on the upper copy and $-F(-x)$ on the lower copy. The boundary condition makes these definitions agree on the seam, and the resulting map $S^n\to S^{n-1}$ is antipodal. Thus (ii) implies (iii).

Conversely, an antipodal map $g:S^n\to S^{n-1}$ restricted to a closed hemisphere, identified with $B^n$, is antipodal on its equatorial boundary. Hence (iii) implies (ii). The three assertions are equivalent; they are standard forms of the Borsuk-Ulam theorem.

For $v\in S^{n-1}$, orthogonally project every member of ${\mathcal{C}}_i$ onto the oriented line ${\ell }_v$. These projections are compact intervals. They are pairwise intersecting because the original sets are, and pairwise intersecting intervals have a common intersection. Let $m_i(v)$ be the midpoint of that common interval and let $f_i(v)$ be its signed coordinate on ${\ell }_v$. Compactness makes $f_i$ continuous, and reversing the orientation gives

Define the continuous antipodal mapThe Borsuk-Ulam theorem gives $v$ with $F(v)=0$. Write the common value as $t$. The hyperplanemeets every set in every ${\mathcal{C}}_i$, because $t=f_i(v)$ lies in the projection interval of each such set. Thus $H$ is the required common hyperplane transversal.