Regularization and angular control

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Let \(P\), \(Q\), \(T=P+Q\), \(\lambda\) and \(\mu\) be as in Proposition 5.1 of Taming, compatibility and a separating current. This lesson proves that the correction \(Q\) is square integrable, and that nearby complex lines of \(P\) are close on average: the angular integral \[ \mathcal I_r=\iint_{d(x,y)<r}|\ell-\tau_{yx}m|^2\,d\lambda\,d\lambda \] is at most \(Cr^4\); here \((x,\ell)\) and \((y,m)\) both range over \(\mathscr C\) with the measure \(\lambda\), and \(\tau_{yx}\) is parallel transport along the shortest geodesic from \(y\) to \(x\). Both statements come from one identity. A closed current paired with the Hodge star of a regularization of another closed current gives zero; for \(T\) paired with itself, the positive part \(P\) contributes the angular integral with a positive weight, up to an error that quadratic mass growth controls.

The lesson follows [OpenAI-TC, Section 4]. Elliptic operators are those of Symbols, finite defects, and the index on a closed manifold (course Elliptic boundary reduction), and \(\rho_0\) and normal coordinates are as in Quadratic mass growth. All forms are real, \(\langle\cdot,\cdot\rangle\) and \(\|\cdot\|\) denote the \(L^2\) product and norm unless a point is indicated, and \(0<r\le1\).

1. The Hodge Laplacian and its resolvent powers

Let \(\Delta=dd^*+d^*d\) act on forms of every degree.

Lemma 1.1. (1) \(\Delta\) is formally self-adjoint, \(\langle\Delta\alpha,\alpha\rangle=\|d\alpha\|^2+\|d^*\alpha\|^2\), and \(\Delta\) is elliptic of order two with principal symbol \(|\xi|^2\). It commutes with \(d\) and \(d^*\), and with \(*\) on two-forms. (2) The square integrable two-forms have an orthonormal basis of smooth eigenforms, \(\Delta\varphi_j=\lambda_j\varphi_j\), with \(\lambda_j\ge0\) and \(\lambda_j\to\infty\). (3) For every integer \(k\ge0\), \(\|u\|_{H^{2k}}\) and \(\|(1+\Delta)^ku\|\) are equivalent norms on \(H^{2k}\). (4) For every real \(s\) and every \(v\in H^{s+1}\), \(\|v\|_{H^s}^2\le C_s\|v\|_{H^{s-1}}\|v\|_{H^{s+1}}\).

Proof. (1) The first two claims are integration by parts. The symbols of \(d\) and \(d^*\) are \(i\xi\wedge\) and \(-i\iota_\xi\), and \(\xi\wedge\iota_\xi+\iota_\xi\,\xi\wedge=|\xi|^2\). Next, \(d\Delta=dd^*d=\Delta d\), and likewise for \(d^*\). In dimension four \(d^*=-*d*\) in every degree, and a direct expansion, using \(**=1\) on two-forms and \(**=-1\) on one- and three-forms, gives \(*\Delta=\Delta*\) on two-forms.

(2) By (1), \(1+\Delta\) is elliptic, and it is injective on every Sobolev space: by (I5) of the lesson cited above a solution of \((1+\Delta)u=0\) is smooth, and then \(\|u\|^2+\|du\|^2+\|d^*u\|^2=0\). Being formally self-adjoint, it is surjective by (I7). So \(K=(1+\Delta)^{-1}\) maps \(L^2\) into \(H^2\); it is compact on \(L^2\) by (I2), self-adjoint, positive and injective. By Theorem 6.2 of Hilbert spaces and compact operators (course Foundations of von Neumann algebras), \(L^2\) has an orthonormal basis of eigenvectors of \(K\) with eigenvalues \(\kappa_j\to0\), \(\kappa_j>0\). Then \(\varphi_j\) is smooth by elliptic regularity, and \(\lambda_j=\kappa_j^{-1}-1\ge0\).

(3) \((1+\Delta)^k\) is elliptic of order \(2k\) and injective as in (2), hence an isomorphism \(H^{2k}\to L^2\) with bounded inverse.

(4) On \(\mathbb R^4\), Cauchy–Schwarz gives \[ \begin{aligned} \int\langle\xi\rangle^{2s}|\hat v|^2&\le\Bigl(\int\langle\xi\rangle^{2s-2}|\hat v|^2\Bigr)^{1/2}\\ &\quad\times\Bigl(\int\langle\xi\rangle^{2s+2}|\hat v|^2\Bigr)^{1/2} . \end{aligned} \] The norm of \(H^s(X)\) is equivalent to the square root of the sum of the squared local norms of \(\chi_iv\) for a fixed partition (Section 1 of the lesson on the index on a closed manifold), and Cauchy–Schwarz for the finite sum gives the claim. \(\blacksquare\)

For a bounded Borel function \(f\) on \([0,\infty)\) put \(f(\Delta)u=\sum_jf(\lambda_j)\langle u,\varphi_j\rangle\varphi_j\) on square integrable two-forms. In particular \[ S_r=(1+r^2\Delta)^{-3},\qquad e^{-\varepsilon\Delta}\quad(\varepsilon>0) . \]

Lemma 1.2 (resolvent powers). For \(0<r\le1\):

  1. \(S_r\) is self-adjoint on \(L^2\), \(\|S_r\|\le1\), \(S_r*=*S_r\), and \(dS_r=S_rd\) and \(d^*S_r=S_rd^*\) on smooth forms of all degrees.
  2. \(A_r=(1+r^2\Delta)^3\) is an isomorphism \(H^{s}\to H^{s-6}\) for every \(s\), and \(S_r\) is the restriction of its inverse to \(L^2\). In particular \(S_r\) extends to a continuous map \(H^{-4}\to H^2\), self-adjoint for the duality, with \((S_rW)(\alpha)=W(S_r\alpha)\).
  3. \(\|S_ru\|_{H^k}\le C_kr^{-k}\|u\|\) for \(u\in L^2\) and \(0\le k\le6\).
  4. For smooth \(\alpha\), \(S_r\alpha\to\alpha\) in every \(H^s\) as \(r\to0\); for \(u\in L^2\), \(S_ru\to u\) in \(L^2\).
  5. \(e^{-\varepsilon\Delta}\) maps \(L^2\) into smooth forms, \(e^{-\varepsilon\Delta}u\to u\) in \(L^2\) as \(\varepsilon\to0\), and \(\langle e^{-\varepsilon\Delta}u,d\gamma\rangle=\langle u,d\,e^{-\varepsilon\Delta}\gamma\rangle\) for smooth one-forms \(\gamma\).

Proof. (1) The operator \(f(\Delta)\) is self-adjoint with norm \(\sup|f|\le1\). Since \(*\) maps eigenforms to eigenforms with the same eigenvalue, it commutes with \(f(\Delta)\). For smooth \(\alpha\), the expansion \(\sum\langle\alpha,\varphi_j\rangle\varphi_j\) converges in every \(H^{2k}\) by Lemma 1.1(3), and \(d\varphi_j\) is either zero or an eigenform with eigenvalue \(\lambda_j\); this gives \(dS_r=S_rd\), and similarly for \(d^*\).

(2) \(A_r\) is elliptic of order six and injective, so it is an isomorphism on the Sobolev scale. On \(L^2\) it acts diagonally, so \(A_r^{-1}=S_r\) there, and the duality statement follows from the formal self-adjointness of \(A_r\).

(3) For \(k=2i\), Lemma 1.1(3) and \(\sup_{\lambda\ge0}\lambda^j(1+r^2\lambda)^{-3}\le r^{-2j}\) for \(j\le3\) give \[ \begin{aligned} \|S_ru\|_{H^{2i}}&\le C\sum_{j\le i}\binom ij\|\Delta^jS_ru\|\\ &\le Cr^{-2i}\|u\| . \end{aligned} \] For odd \(k=2i-1\), Lemma 1.1(4) interpolates between \(2i-2\) and \(2i\).

(4) By dominated convergence in the expansion, since \(\|(S_r-1)\alpha\|_{H^{2k}}\) is at most \(C\|(S_r-1)(1+\Delta)^k\alpha\|\).

(5) \(\Delta^ke^{-\varepsilon\Delta}u\in L^2\) for all \(k\), so \(e^{-\varepsilon\Delta}u\) lies in every \(H^{2k}\), and is smooth by (I1). Convergence is dominated convergence. The last identity holds because \(e^{-\varepsilon\Delta}\) is self-adjoint and commutes with \(d\) on smooth forms. \(\blacksquare\)

The currents \(P\), \(Q\), \(T\) lie in \(H^{-3}\subseteq H^{-4}\), so \(S_rP\), \(S_rQ\), \(S_rT\) are square integrable forms. The same holds for every subcurrent \[ P'(\alpha)=\int_{\mathscr C}\alpha_y(m)\,d\lambda'(y,m), \] where \(\lambda'\le\lambda\) is a Borel measure on \(\mathscr C\); its projection \(\mu'\) satisfies \(\mu'\le\mu\), and \(|P'(\alpha)|\le\|\alpha\|_\infty\).

2. The zero pairing

Lemma 2.1 ([OpenAI-TC, Lemma 4.1]). If \(T'\) is a closed current in \(H^{-4}\), then \[ \langle S_rT,\,*S_rT'\rangle=0 . \]

Proof. \(S_rT\) annihilates every smooth closed two-form \(\alpha\): \(S_r\alpha\) is smooth and closed by Lemma 1.2(1), and \((S_rT)(\alpha)=T(S_r\alpha)=0\). Let \(w=S_rT'\in L^2\). For a smooth one-form \(\gamma\), \(\langle w,d\gamma\rangle=T'(S_rd\gamma)=T'(dS_r\gamma)=0\). By Lemma 1.2(5), \(w_\varepsilon=e^{-\varepsilon\Delta}w\) is smooth with \(\langle w_\varepsilon,d\gamma\rangle=0\) for all \(\gamma\), so \(d^*w_\varepsilon=0\). In dimension four \(*d^*=d*\) on two-forms, so \(*w_\varepsilon\) is a smooth closed form. Hence \(\langle S_rT,*w_\varepsilon\rangle=0\), and \(*w_\varepsilon\to*w\) in \(L^2\). \(\blacksquare\)

3. A heat kernel hypothesis and the kernel of the squared resolvent power

We use one fact about the heat kernel that the programme does not prove. Let \(\chi_1\) be a fixed smooth function on \([0,\infty)\), equal to one on \([0,\rho_0]\) and zero on \([2\rho_0,\infty)\).

Hypothesis (HK). For \(t>0\) the operator \(e^{-t\Delta}\) on two-forms has a kernel \(k_t(x,y)\colon\Lambda^2_y\to\Lambda^2_x\), continuous in \((t,x,y)\). There are \(t_0>0\), \(c_0>0\), \(C\), and a smooth positive function \(a\) on \(\{d(x,y)<2\rho_0\}\), such that, with \(\tau_{yx}\colon\Lambda^2_y\to\Lambda^2_x\) the parallel transport of the Levi-Civita connection along the shortest geodesic, for \(0<t<t_0\) and every integer \(M\ge0\), \[ \begin{aligned} &\Bigl|k_t(x,y)-\chi_1(d)\frac{e^{-d^2/(4t)}}{(4\pi t)^2}a(x,y)\tau_{yx}\Bigr|\\ &\qquad\le Ct^{-1}e^{-c_0d^2/t}+C_Mt^M , \end{aligned} \] where \(d=d(x,y)\); and \(|k_t(x,y)|\le C\) for \(t\ge t_0\).

This is the leading term of the short-time expansion of the heat kernel of a Laplace-type operator, with the strong remainder estimate of Ludewig [Ludewig 2019, Theorem 1.1] near the diagonal and his Gaussian upper bound [Ludewig 2019, Theorem 3.5] away from it; the leading coefficient is parallel transport times \(J(x,y)^{-1/2}\), with \(J\) the Jacobian of the exponential map [Ludewig 2018, Section 3, (3.4)]. Ludewig's transport is that of the connection for which the operator is the connection Laplacian plus an endomorphism; for the Hodge Laplacian this is the Levi-Civita connection, by the Weitzenböck formula. Neither the expansion nor the Weitzenböck formula is proved in the programme, so (HK) is a named hypothesis of this course. The rest of the argument is proved here.

Lemma 3.1 (the kernel of \(S_r^2\); [OpenAI-TC, Lemma 4.2]). Assume (HK). There is \(r_0\in(0,1]\) such that for \(0<r\le r_0\) the operator \(S_r^2\) has a continuous kernel \[ K_r(x,y)=p_r(x,y)\tau_{yx}+E_r(x,y) , \] where \(p_r\ge0\) vanishes where \(d(x,y)\ge2\rho_0\) (and \(p_r\tau_{yx}\) is read as zero there), and for every integer \(N\ge0\), with \(d=d(x,y)\), \[ \begin{gathered} p_r\le C_Nr^{-4}(1+d/r)^{-N},\\ p_r\ge cr^{-4}\quad\text{where }d<r,\\ |E_r|\le C_Nr^{-2}(1+d/r)^{-N} . \end{gathered} \]

Proof. For \(\lambda\ge0\), \(\int_0^\infty s^5e^{-s(1+r^2\lambda)}\,ds=\Gamma(6)(1+r^2\lambda)^{-6}\). Since \(\|e^{-t\Delta}\|\le1\), this gives \[ S_r^2=\frac1{\Gamma(6)}\int_0^\infty e^{-s}s^5e^{-r^2s\Delta}\,ds \] as a norm-convergent integral of operators. The bounds of (HK) make \(\int_0^\infty e^{-s}s^5k_{r^2s}(x,y)\,ds\) absolutely and locally uniformly convergent, and testing against smooth forms with Fubini's theorem shows that it is \(\Gamma(6)K_r\).

Let \(r_0^2\le t_0/2\) and \(r_0\le\rho_0\). Define \(p_r\) as \(\Gamma(6)^{-1}\) times the integral of the leading term over \(r^2s<t_0\): \[ \begin{aligned} p_r&=\frac{\chi_1(d)a}{\Gamma(6)(4\pi r^2)^2}\\ &\quad\times\int_0^{t_0/r^2}e^{-s}s^3e^{-d^2/(4r^2s)}\,ds . \end{aligned} \] It is nonnegative. For every \(N\), \(e^{-d^2/(4r^2s)}\le C_N(1+\sqrt s)^N(1+d/r)^{-N}\). Indeed, if \(d/r\le\sqrt s\) the right side is at least \(C_N\ge1\); otherwise, with \(w=d/(r\sqrt s)>1\), \((1+d/r)\le(1+\sqrt s)(1+w)\) and \(e^{-w^2/4}\le C_N(1+w)^{-N}\). Since \(\int_0^\infty s^3(1+\sqrt s)^Ne^{-s}\,ds<\infty\), the upper bound for \(p_r\) follows. If \(d<r\), the part \(1\le s\le2\) of the integral (inside the range, as \(2r^2\le t_0\)) has \(e^{-d^2/(4r^2s)}\ge e^{-1/4}\) and \(\chi_1(d)=1\), and \(a\) is bounded below; this gives \(p_r\ge cr^{-4}\).

The rest of \(K_r\) is \(E_r\). For \(r^2s<t_0\), the remainder in (HK) contributes \(\Gamma(6)^{-1}\) times the integral of \(e^{-s}s^5\) against \(C(r^2s)^{-1}e^{-c_0d^2/(r^2s)}+C_M(r^2s)^M\). The first term is at most \(C_Nr^{-2}(1+d/r)^{-N}\) by the same inequality, with \(c_0\) in place of \(\frac14\). The second is \(O(r^{2M})\); as \(d\) is at most the diameter \(D\) of \(X\), \(r^{2M}\le C r^{-2}(1+d/r)^{-N}\) once \(2M\ge N-2\). For \(s\ge t_0/r^2\), the kernel is bounded and \(\int_{t_0/r^2}^\infty e^{-s}s^5\,ds\le Ce^{-t_0/(2r^2)}\), again \(O(r^{-2}(1+d/r)^{-N})\). \(\blacksquare\)

4. The angular identity and the positive pairing

Lemma 4.1 (the angular identity; [OpenAI-TC, (4.5)]). Let \(d(x,y)<2\rho_0\), \(\ell\in\mathscr C_x\) and \(m\in\mathscr C_y\). Then \[ \begin{aligned} &\langle\ell,*\tau_{yx}m\rangle\\ &\quad=\tfrac12\langle F_x,\tau_{yx}F_y\rangle-\langle\ell,\tau_{yx}m\rangle\\ &\quad=\tfrac12|\ell-\tau_{yx}m|^2-\tfrac14|F_x-\tau_{yx}F_y|^2 , \end{aligned} \] and \(|F_x-\tau_{yx}F_y|\le C\,d(x,y)\).

Proof. Levi-Civita parallel transport is an orientation-preserving isometry, so on two-forms it is an isometry commuting with \(*\). By Lemma 1.1(3) of the first lesson, \(*\ell=F_x-\ell\), and \(m^+=F_y/2\), so \(\tau_{yx}m^+=\frac12\tau_{yx}F_y\) is self-dual while \(\tau_{yx}m^-\) is anti-self-dual. Hence \[ \begin{aligned} &\langle\ell,*\tau_{yx}m\rangle=\langle F_x-\ell,\tau_{yx}m\rangle\\ &\quad=\tfrac12\langle F_x,\tau_{yx}F_y\rangle-\langle\ell,\tau_{yx}m\rangle . \end{aligned} \] The second form follows from \(|\ell|=|m|=1\) and \(|F|^2=2\). The map \((x,y)\mapsto\tau_{yx}F_y-F_x\) is smooth on \(\{d<2\rho_0\}\), by smooth dependence of geodesics and of parallel transport on parameters (Connections and parallel transport; Theorem B.5 of Riemannian connections and convex neighbourhoods), and vanishes on the diagonal. \(\blacksquare\)

The identity is the heart of the method. Pairing \(P\) with the Hodge star of its own regularization does not measure mass; it measures how far apart nearby complex lines are, up to the drift of \(F\) under transport, which is of second order in the distance.

Lemma 4.2 (the positive pairing; [OpenAI-TC, Lemma 4.4]). Assume (HK) and \(0<r\le r_0\). Let \(P'\) be a subcurrent with measure \(\lambda'\le\lambda\) and projection \(\mu'\). (1) For every \(y\in X\), \(r^{-2}\int_X(1+d(x,y)/r)^{-6}\,d\mu(x)\le C\). (2) \(\|S_rP'\|\le Cr^{-1}\). (3) With \(d=d(x,y)\), and with \((x,\ell)\) distributed by \(\lambda\) and \((y,m)\) by \(\lambda'\), \[ \begin{aligned} &\langle S_rP,*S_rP'\rangle\\ &\quad\ge cr^{-4}\iint_{d<r}|\ell-\tau_{yx}m|^2\,d\lambda\,d\lambda'\\ &\qquad-Cr^{-2}\iint(1+d/r)^{-6}\,d\mu\,d\mu' . \end{aligned} \]

Proof. (1) is the dyadic bound after Proposition 4.1 of Quadratic mass growth.

For the pairings, \(S_r^2\) maps \(H^{-4}\) into \(H^8\), which consists of continuous forms, and testing against smooth forms shows \((S_r^2P')(x)=\int K_r(x,y)m\,d\lambda'(y,m)\), using \(K_r(x,y)^*=K_r(y,x)\). Hence, since \(S_r\) is self-adjoint and commutes with \(*\), \[ \begin{aligned} &\langle S_rP,*S_rP'\rangle=P(*S_r^2P')\\ &\quad=\iint\langle\ell,*K_r(x,y)m\rangle\,d\lambda\,d\lambda' . \end{aligned} \] Insert \(K_r=p_r\tau_{yx}+E_r\) and Lemma 4.1. The term \(\frac12p_r|\ell-\tau_{yx}m|^2\) is nonnegative; keep it only where \(d<r\), where \(p_r\ge cr^{-4}\). The term \(-\frac14p_r|F_x-\tau_{yx}F_y|^2\) is at least \(-C_Nr^{-4}(1+d/r)^{-N}d^2\), which for \(N=8\) is at least \(-Cr^{-2}(1+d/r)^{-6}\). The error \(\langle\ell,*E_rm\rangle\) is at least \(-|E_r|\ge-Cr^{-2}(1+d/r)^{-6}\). Integrating over the fibers, whose measures project to \(\mu\) and \(\mu'\), gives (3).

For (2), with \((x,m')\) and \((y,m)\) distributed by \(\lambda'\), \[ \begin{aligned} \|S_rP'\|^2&=P'(S_r^2P')\\ &=\iint\langle m',K_r(x,y)m\rangle\,d\lambda'\,d\lambda' , \end{aligned} \] and \(|K_r|\le Cr^{-4}(1+d/r)^{-6}\). With \(g_r(y)=r^{-2}\int(1+d(x,y)/r)^{-6}d\mu'(x)\), which is at most \(C\) by (1) and \(\mu'\le\mu\), \[ \|S_rP'\|^2\le Cr^{-2}\int g_r\,d\mu'\le Cr^{-2} . \] \(\blacksquare\)

5. Mixing of invariant and anti-invariant parts

Lemma 5.1 (relative projection estimate; [OpenAI-TC, Lemma 4.5]). Let \(\Pi\) be a smooth field of orthogonal projections on \(\Lambda^2T^*X\). If \(W\in H^{-4}\) satisfies \(\Pi W=0\), then \[ \|\Pi S_rW\|\le Cr\,\|S_rW\| , \] with \(C\) depending on \(\Pi\) but not on \(r\) or \(W\).

Proof. Let \(U=S_rW\in H^2\). By Lemma 1.2(2), \(A_rU=W\) and \(\Pi U=S_rA_r\Pi U\). Since \(\Pi W=0\), \[ A_r\Pi U=\Pi A_rU+[A_r,\Pi]U=[A_r,\Pi]U , \] so \(\Pi U=S_r[A_r,\Pi]U\). Expanding, \([A_r,\Pi]=\sum_{j=1}^3\binom3jr^{2j}[\Delta^j,\Pi]\). The principal symbol of \(\Delta^j\) is the scalar \(|\xi|^{2j}\), so \([\Delta^j,\Pi]\) is a differential operator of order at most \(2j-1\), bounded \(H^{2j-1}\to L^2\). By Lemma 1.2(3), for the operator norm on \(L^2\), \[ \|r^{2j}[\Delta^j,\Pi]S_r\|\le Cr^{2j}r^{-(2j-1)}=Cr . \] For smooth \(u\) and \(v\), \(\langle S_r[A_r,\Pi]u,v\rangle=-\langle u,[A_r,\Pi]S_rv\rangle\), because \(A_r\), \(\Pi\) and \(S_r\) are formally self-adjoint. So \(S_r[A_r,\Pi]\) extends from smooth forms to a bounded operator on \(L^2\) of norm at most \(Cr\), and the extension agrees with its action on \(U\), as both are continuous from \(L^2\) to \(H^{-6}\). Hence \(\|\Pi U\|\le Cr\|U\|\). \(\blacksquare\)

Thus regularization at scale \(r\) can create only an \(O(r)\) fraction of a component that the current did not have. For \(\Pi=R\) and \(W=P'\) (note \(RP'=0\), since every \(m\in\mathscr C\) is invariant), and for \(\Pi=1-R\) and an anti-invariant \(W\), this controls the cross terms between \(P\) and \(Q\).

6. The correction is square integrable

Proposition 6.1 ([OpenAI-TC, Proposition 4.6]). Assume (HK). Then \(Q\in L^2\), and \[ \mathcal I_r\le Cr^4 \] for \(0<r\le r_0\). For every subcurrent \(P'\), \[ \lim_{r\to0}\langle S_rP',\,*S_rQ\rangle=0 . \]

Proof. Cross terms. Let \(V\in H^{-4}\) be anti-invariant, that is \((1-R)V=0\). Since \(R\) is a pointwise orthogonal projection and \(RP'=0\), \[ \begin{aligned} \langle S_rP',S_rV\rangle&=\langle RS_rP',S_rV\rangle\\ &\quad+\langle S_rP',(1-R)S_rV\rangle . \end{aligned} \] Lemma 5.1, applied twice, and Lemma 4.2(2) give \[ \begin{aligned} |\langle S_rP',S_rV\rangle|&\le Cr\|S_rP'\|\,\|S_rV\|\\ &\le C\|S_rV\| . \end{aligned} \]

Energy. \(Q\) is self-dual, so \(*S_rQ=S_rQ\), and Lemma 2.1 with \(T'=T\) gives \[ \begin{aligned} 0&=\langle S_rP,*S_rP\rangle\\ &\quad+2\langle S_rP,S_rQ\rangle+\|S_rQ\|^2 . \end{aligned} \] By Lemma 4.2(3) and (1), the first term is at least \(cr^{-4}\mathcal I_r-C\); by the cross-term bound the second is at least \(-C\|S_rQ\|\). Hence \[ cr^{-4}\mathcal I_r+\|S_rQ\|^2\le C+C\|S_rQ\| . \] So \(\|S_rQ\|\) is bounded as \(r\to0\), and \(\mathcal I_r\le Cr^4\). A sequence \(r_k\to0\) has \(S_{r_k}Q\) converging weakly in \(L^2\) to some \(q\): bounded sequences in a Hilbert space have weakly convergent subsequences (Exercise 4 of Weak topologies: Tychonoff, Banach–Alaoglu, Mazur, bipolars, Krein–Milman and Eberlein–Šmulian, with its solution; Hilbert spaces are reflexive by Theorem 2.3 of the lesson on Hilbert spaces cited above). For smooth \(\alpha\), \(\langle S_rQ,\alpha\rangle=Q(S_r\alpha)\to Q(\alpha)\) by Lemma 1.2(4), so \(Q(\alpha)=\langle q,\alpha\rangle\): \(Q=q\in L^2\).

Vanishing cross terms. For anti-invariant \(V\in L^2\) the cross-term bound and \(\|S_r\|\le1\) give \(|\langle S_rP',S_rV\rangle|\le C\|V\|\), uniformly in \(r\). If \(V\) is smooth and anti-invariant, then \(\langle S_rP',S_rV\rangle=P'(S_r^2V)\to P'(V)\) by Lemma 1.2(4), and \(P'(V)=\int\langle V_y,m\rangle\,d\lambda'=0\) since \(V\) is anti-invariant and \(m\) invariant. Smooth anti-invariant forms are dense among square integrable anti-invariant forms (approximate by smooth forms, then apply \(R\)). Approximating \(Q\) in \(L^2\) by such \(V\) gives \(\langle S_rP',S_rQ\rangle\to0\), and \(*S_rQ=S_rQ\). \(\blacksquare\)

Two consequences are used in the last lesson: \(S_rQ\to Q\) in \(L^2\) (Lemma 1.2(4)), and the pairing of \(P\) with a subcurrent \(P'\) has the lower bound of Lemma 4.2(3), whose negative term will be shown to vanish when \(\mu'\) lives on points of zero density.

7. Exercises

7.1. Show that \(\sup_{\lambda\ge0}\lambda^j(1+r^2\lambda)^{-3}\) equals \(\bigl(\frac{j}{3-j}\bigr)^j\frac{(3-j)^3}{27}\,r^{-2j}\) for \(j=1,2\), and that it is at most \(r^{-2j}\) for \(0\le j\le3\).

7.2. Let \(\ell\) and \(m\) be unit complex-line bivectors at the same point with \(F\) the same. Show that \(\langle\ell,*m\rangle=\frac12|\ell-m|^2\ge0\), with equality only for \(\ell=m\).

7.3. On \(\mathbb R^4\) with the Euclidean metric, the heat kernel on functions is \((4\pi t)^{-2}e^{-|x-y|^2/(4t)}\). Show that the kernel of \((1-r^2\Delta_0)^{-6}\), with \(\Delta_0=\sum_i\partial_i^2\), is \(r^{-4}\Phi(|x-y|/r)\) for a positive function \(\Phi\), and give \(\Phi\) as an integral.

7.4. On the flat torus \(\mathbb R^4/\mathbb Z^4\), write two-forms in the constant frame \(dx^i\wedge dx^j\), and let \(\Pi\) be a constant orthogonal projection in this frame. Show that \(\Pi S_r=S_r\Pi\), so that Lemma 5.1 holds with \(C=0\).

7.5. Show that \(\langle\alpha,*\beta\rangle=\int_X\alpha\wedge\beta\) for square integrable two-forms. On the flat torus \(\mathbb R^4/\mathbb Z^4\), let \(T\) and \(T'\) be the currents given by the constant forms \(dx^1\wedge dx^2\) and \(dx^3\wedge dx^4\). Show that both are closed and that \(\langle S_rT,*S_rT'\rangle=1\). Which hypothesis of Lemma 2.1 fails?

8. Solutions

7.1. For \(f(\lambda)=\lambda^j(1+r^2\lambda)^{-3}\), \(f'=0\) when \(j(1+r^2\lambda)=3r^2\lambda\), that is \(r^2\lambda=j/(3-j)\). Then \(1+r^2\lambda=3/(3-j)\), and \(f=r^{-2j}(j/(3-j))^j((3-j)/3)^3\). For \(j=1\) this is \(\frac4{27}r^{-2}\), for \(j=2\) it is \(\frac4{27}r^{-4}\). For \(j=3\), \(\lambda^3(1+r^2\lambda)^{-3}<r^{-6}\), and for \(j=0\) the supremum is \(1\).

7.2. With \(x=y\) and \(\tau\) the identity, Lemma 4.1 gives \(\langle\ell,*m\rangle=\frac12|\ell-m|^2-\frac14|F-F|^2\).

7.3. The integral formula of the proof of Lemma 3.1, with \(e^{t\Delta_0}\) in place of \(e^{-t\Delta}\), gives the kernel \[ \frac1{\Gamma(6)}\int_0^\infty e^{-s}s^5\frac{e^{-|x-y|^2/(4r^2s)}}{(4\pi r^2s)^2}\,ds , \] which is \(r^{-4}\Phi(|x-y|/r)\) with \(\Phi(\rho)=\frac{1}{16\pi^2\Gamma(6)}\int_0^\infty e^{-s}s^3e^{-\rho^2/(4s)}\,ds>0\).

7.4. On the flat torus, \(\Delta\) acts componentwise as \(-\sum_i\partial_i^2\) in the constant frame, and a constant \(\Pi\) commutes with it, hence with \(S_r\). Then \(\Pi W=0\) gives \(\Pi S_rW=S_r\Pi W=0\).

7.5. Since \(\alpha\wedge*\gamma=\langle\alpha,\gamma\rangle\,dV\) and \(**=1\) on two-forms, \(\langle\alpha,*\beta\rangle=\int\alpha\wedge**\beta=\int\alpha\wedge\beta\). Constant forms on the flat torus satisfy \(d^*=0\), so \(T(d\gamma)=\langle d^*dx^{12},\gamma\rangle=0\): both currents are closed. Constant forms are harmonic, so \(S_r\) fixes them, and \(\langle S_rT,*S_rT'\rangle=\int dx^1\wedge dx^2\wedge dx^3\wedge dx^4\), which is \(1\). Lemma 2.1 needs \(T\) to annihilate every closed form, and \(T(dx^1\wedge dx^2)=1\). In this language the lemma says that the intersection pairing \(\int S_rT\wedge S_rT'\) vanishes when \(T\) annihilates all closed forms.

References