[proofplan] The proof uses the standard Schoen-Uhlenbeck partial regularity scheme for minimizing harmonic maps. First define the regular set by small-scale energy decay and use epsilon regularity to prove smoothness there; the complement is therefore relatively closed. Then monotonicity and compactness produce homogeneous minimizing tangent maps at singular points. Finally, dimension reduction excludes singular strata of codimension less than $3$, which gives the Hausdorff dimension estimate and hence the weaker $\mathcal H^{m-2}$ vanishing statement. [/proofplan]

[step:Define the regular set by small-scale normalized energy]Let $q\in\mathbb N$ be such that $N$ is isometrically embedded in $\mathbb R^q$, and regard $u$ as a map $u:U\to\mathbb R^q$ with $u(x)\in N$ for $\mathcal L^m$-a.e. $x\in U$. Here $\mathcal L^m$ denotes $m$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R^m$. Since $U$ is open, for every $a\in U$ there exists $r_a>0$ with $B(a,r_a)\subset U$; this is the interior-ball property used below. For $a\in U$ and $r>0$ with $B(a,r)\subset U$, define the normalized Dirichlet energy $\Theta_u(a,r)\in[0,\infty)$ by \begin{align*} \Theta_u(a,r):=r^{2-m}\int_{B(a,r)} |\nabla u(x)|^2\,d\mathcal L^m(x). \end{align*} Here $\nabla u:U\to\mathbb R^{q\times m}$ denotes the weak gradient of the Euclidean representative of $u$. We use the [Schoen-Uhlenbeck epsilon regularity theorem for minimizing harmonic maps](/page/Schoen-Uhlenbeck%20Epsilon%20Regularity%20Theorem): there exist constants $\varepsilon_0=\varepsilon_0(m,N)>0$ and $\alpha=\alpha(m,N)\in(0,1)$ such that, whenever $B(a,2r)\subset U$ and $u$ minimizes the Dirichlet energy against compactly supported competitors in $B(a,2r)$, the bound $\Theta_u(a,2r)<\varepsilon_0$ implies $u\in C^\infty(B(a,r);N)$. Define the regular set $\operatorname{reg}(u)\subset U$ by \begin{align*} \operatorname{reg}(u):=\{a\in U: \text{there exists } r>0 \text{ with } B(a,2r)\subset U \text{ and } \Theta_u(a,2r)<\varepsilon_0\}. \end{align*} Define the singular set $\operatorname{sing}(u)\subset U$ by \begin{align*} \operatorname{sing}(u):=U\setminus\operatorname{reg}(u). \end{align*} By epsilon regularity, $u$ is smooth on every ball $B(a,r)$ with $a\in\operatorname{reg}(u)$ chosen as above. Hence $u$ is smooth on $\operatorname{reg}(u)$.[/step]

[guided]The first task is to turn the analytic regularity criterion into a set-theoretic decomposition of $U$. Since $u\in W^{1,2}(U;N)$ and $N\subset\mathbb R^q$ is isometrically embedded, we may treat $u$ as a Euclidean Sobolev map $u:U\to\mathbb R^q$ whose values lie in $N$ almost everywhere. Its weak gradient $\nabla u:U\to\mathbb R^{q\times m}$ is square-integrable on compact subsets of $U$. For $a\in U$ and $r>0$ satisfying $B(a,r)\subset U$, define \begin{align*} \Theta_u(a,r):=r^{2-m}\int_{B(a,r)} |\nabla u(x)|^2\,d\mathcal L^m(x). \end{align*} The factor $r^{2-m}$ is chosen because the Dirichlet energy rescales like $r^{m-2}$ under the blow-up $x=a+ry$. Thus $\Theta_u(a,r)$ is the scale-invariant quantity controlled by the monotonicity and epsilon regularity theory. We now invoke the [Schoen-Uhlenbeck epsilon regularity theorem for minimizing harmonic maps](/page/Schoen-Uhlenbeck%20Epsilon%20Regularity%20Theorem). Its hypotheses are satisfied on every ball $B(a,2r)\subset U$: the theorem statement assumes that $u$ is energy-minimizing, so its restriction to $B(a,2r)$ minimizes the Dirichlet energy among competitors agreeing with $u$ outside a compact subset of that ball; the target $N$ is compact and smooth; and the domain ball is Euclidean. Therefore there are constants $\varepsilon_0=\varepsilon_0(m,N)>0$ and $\alpha=\alpha(m,N)\in(0,1)$ such that \begin{align*} \Theta_u(a,2r)<\varepsilon_0 \quad\Longrightarrow\quad u\in C^\infty(B(a,r);N). \end{align*} This motivates the definition \begin{align*} \operatorname{reg}(u):=\{a\in U: \text{there exists } r>0 \text{ with } B(a,2r)\subset U \text{ and } \Theta_u(a,2r)<\varepsilon_0\}. \end{align*} We then define \begin{align*} \operatorname{sing}(u):=U\setminus\operatorname{reg}(u). \end{align*} For each $a\in\operatorname{reg}(u)$, the defining radius $r$ gives smoothness on $B(a,r)$ by epsilon regularity. Since these balls cover $\operatorname{reg}(u)$, the map $u$ is smooth throughout $\operatorname{reg}(u)$.[/guided]

[step:Show the singular set is relatively closed] We prove that $\operatorname{reg}(u)$ is open in the relative topology of $U$. Let $a\in\operatorname{reg}(u)$, and choose $r>0$ such that $B(a,2r)\subset U$ and $\Theta_u(a,2r)<\varepsilon_0$. Since $B(a,r/2)\subset B(a,r)$ and epsilon regularity gives $u\in C^\infty(B(a,r);N)$, the function $g:B(a,r)\to[0,\infty)$ defined by $g(x):=|\nabla u(x)|^2$ is continuous. Fix $b\in B(a,r/4)$. Choose $\rho_b>0$ so small that $B(b,2\rho_b)\subset B(a,r)$; this is possible because $B(a,r)$ is open and $b\in B(a,r/4)$. Since $g$ is continuous, it is locally bounded near $b$: there exist $M_b>0$ and $\sigma_b>0$ such that $g(x)\le M_b$ for every $x\in B(b,\sigma_b)$. Reducing $\rho_b$ if necessary so that $2\rho_b\le\sigma_b$, we obtain \begin{align*} (2\rho_b)^{2-m}\int_{B(b,2\rho_b)} |\nabla u(x)|^2\,d\mathcal L^m(x) \le (2\rho_b)^{2-m}M_b\mathcal L^m(B(b,2\rho_b)). \end{align*} Since $\mathcal L^m(B(b,2\rho_b))=\mathcal L^m(B(0,1))(2\rho_b)^m$, this becomes \begin{align*} (2\rho_b)^{2-m}\int_{B(b,2\rho_b)} |\nabla u(x)|^2\,d\mathcal L^m(x) \le M_b\mathcal L^m(B(0,1))(2\rho_b)^2. \end{align*} The right-hand side tends to $0$ as $\rho_b\downarrow0$, so we choose $\rho_b$ smaller once more to make it strictly less than $\varepsilon_0$. Thus $b\in\operatorname{reg}(u)$, so $B(a,r/4)\subset\operatorname{reg}(u)$. Therefore $\operatorname{reg}(u)$ is open in $U$, and $\operatorname{sing}(u)=U\setminus\operatorname{reg}(u)$ is relatively closed in $U$. [/step]

[step:Blow up at a singular point to obtain a homogeneous minimizing tangent map]Let $a\in\operatorname{sing}(u)$. For every $r>0$ with $B(a,2r)\subset U$, the definition of $\operatorname{sing}(u)$ gives \begin{align*} \Theta_u(a,2r)\ge \varepsilon_0. \end{align*} The [monotonicity formula for minimizing harmonic maps](/page/Monotonicity%20Formula%20for%20Minimizing%20Harmonic%20Maps) states that $r\mapsto\Theta_u(a,r)$ is nondecreasing on the set of radii with $B(a,r)\subset U$. Hence the density $\Theta_u(a,0^+)\in[\varepsilon_0,\infty)$ defined by \begin{align*} \Theta_u(a,0^+):=\lim_{r\downarrow 0}\Theta_u(a,r) \end{align*} exists. Let $\operatorname{dist}(a,\partial U):=\inf\{|a-z|:z\in\partial U\}$ denote the Euclidean distance from $a$ to the boundary of $U$. Choose a sequence $(r_k)_{k\in\mathbb N}$ with $r_k\downarrow0$. Define the radii $R_k\in(0,\infty]$ by \begin{align*} R_k:=\operatorname{dist}(a,\partial U)/r_k. \end{align*} Define the blow-up maps $u_k:B(0,R_k)\to N$ by \begin{align*} u_k(y):=u(a+r_k y),\qquad y\in B(0,R_k). \end{align*} For every fixed $R>0$, the maps $u_k$ are energy-minimizing on $B(0,R)$ for all sufficiently large $k$, because compactly supported competitors for $u_k$ rescale to compactly supported competitors for $u$ in $B(a,r_kR)\Subset U$. The monotonicity formula gives a uniform $W^{1,2}(B(0,R);\mathbb R^q)$ bound. We invoke the [compactness theorem for minimizing harmonic maps](/page/Compactness%20Theorem%20for%20Minimizing%20Harmonic%20Maps) in the following form: for compact target $N$, a locally energy-bounded sequence of minimizing maps on exhausting Euclidean balls has a subsequence converging strongly in $W^{1,2}_{\mathrm{loc}}$ to a locally minimizing $N$-valued map. Its hypotheses are satisfied by the preceding minimization and energy bounds. Hence, after passing to a subsequence not relabeled, there is a map $\nu_0\in W^{1,2}_{\mathrm{loc}}(\mathbb R^m;N)$ such that $u_k\to\nu_0$ strongly in $W^{1,2}_{\mathrm{loc}}(\mathbb R^m;\mathbb R^q)$. The limiting map $\nu_0$ is energy-minimizing on every ball in $\mathbb R^m$. We now justify homogeneity from the equality case in the [monotonicity formula for minimizing harmonic maps](/page/Monotonicity%20Formula%20for%20Minimizing%20Harmonic%20Maps). For $0<s<t<\infty$, the monotonicity identity for each blow-up $u_k$ gives \begin{align*} \Theta_{u_k}(0,t)-\Theta_{u_k}(0,s) =2\int_{B(0,t)\setminus B(0,s)} |y|^{2-m}\left|\frac{\partial u_k}{\partial r}(y)\right|^2\,d\mathcal L^m(y), \end{align*} where $\partial u_k/\partial r$ denotes the weak radial derivative $\nabla u_k(y)\cdot y/|y|$. The scaling identity gives $\Theta_{u_k}(0,\rho)=\Theta_u(a,r_k\rho)$ for every fixed $\rho>0$ and all sufficiently large $k$. Since $r_k\rho\downarrow0$, the left-hand side converges to \begin{align*} \Theta_u(a,0^+)-\Theta_u(a,0^+)=0. \end{align*} On the fixed annulus $A_{s,t}:=B(0,t)\setminus B(0,s)$, the weight $y\mapsto |y|^{2-m}$ is bounded above and below by positive constants because $s>0$. Strong convergence in $W^{1,2}(A_{s,t};\mathbb R^q)$ gives $\nabla u_k\to\nabla\nu_0$ strongly in $L^2(A_{s,t};\mathbb R^{q\times m})$. Since the radial projection $\xi\mapsto \xi\cdot y/|y|$ has operator norm at most $1$ for each $y\in A_{s,t}$, we have $\partial u_k/\partial r\to\partial\nu_0/\partial r$ strongly in $L^2(A_{s,t};\mathbb R^q)$. Let $h_k:=\partial u_k/\partial r$ and $h:=\partial\nu_0/\partial r$ as elements of $L^2(A_{s,t};\mathbb R^q)$. The elementary inequality $\big||h_k|^2-|h|^2\big|\le |h_k-h|(|h_k|+|h|)$ and the [Cauchy-Schwarz inequality](/theorems/432) give \begin{align*} \int_{A_{s,t}} \big||h_k(y)|^2-|h(y)|^2\big|\,d\mathcal L^m(y) \le \|h_k-h\|_{L^2(A_{s,t})}\bigl(\|h_k\|_{L^2(A_{s,t})}+\|h\|_{L^2(A_{s,t})}\bigr). \end{align*} The right-hand side tends to $0$ because $h_k\to h$ strongly in $L^2(A_{s,t};\mathbb R^q)$ and the sequence $(h_k)_{k\in\mathbb N}$ is therefore bounded in $L^2(A_{s,t};\mathbb R^q)$. Since $|y|^{2-m}$ is bounded on $A_{s,t}$, multiplication by this weight preserves the resulting $L^1$ convergence. Passing to the limit in the weighted radial-energy identity gives \begin{align*} 2\int_{B(0,t)\setminus B(0,s)} |y|^{2-m}\left|\frac{\partial \nu_0}{\partial r}(y)\right|^2\,d\mathcal L^m(y)=0. \end{align*} Therefore $\partial \nu_0/\partial r=0$ for $\mathcal L^m$-a.e. $y\in\mathbb R^m\setminus\{0\}$. Integrating along almost every radial line gives \begin{align*} \nu_0(\lambda y)=\nu_0(y) \end{align*} for every $\lambda>0$ and for $\mathcal L^m$-a.e. $y\in\mathbb R^m$. The tangent map is nonconstant. Indeed, suppose that $\nu_0$ is constant. Strong convergence $u_k\to\nu_0$ in $W^{1,2}(B(0,2);\mathbb R^q)$ gives \begin{align*} \lim_{k\to\infty}\int_{B(0,2)} |\nabla u_k(y)|^2\,d\mathcal L^m(y)=\int_{B(0,2)} |\nabla \nu_0(y)|^2\,d\mathcal L^m(y)=0. \end{align*} Using the scaling identity for the normalized energy at radius $2$, we obtain \begin{align*} \Theta_u(a,2r_k)=\Theta_{u_k}(0,2)=2^{2-m}\int_{B(0,2)} |\nabla u_k(y)|^2\,d\mathcal L^m(y)\to0. \end{align*} Since $2r_k\downarrow0$, the left-hand side also converges to $\Theta_u(a,0^+)$. Hence $\Theta_u(a,0^+)=0$, contradicting $\Theta_u(a,0^+)\ge\varepsilon_0$. Thus every singular point has a nonconstant homogeneous minimizing tangent map.[/step]

[guided]The point of the blow-up is not only to get a limiting minimizer, but to get one with an extra symmetry. We extract that symmetry from the equality case of the [monotonicity formula for minimizing harmonic maps](/page/Monotonicity%20Formula%20for%20Minimizing%20Harmonic%20Maps). For $0<s<t<\infty$, the monotonicity identity applied to $u_k$ on the annulus $B(0,t)\setminus B(0,s)$ says \begin{align*} \Theta_{u_k}(0,t)-\Theta_{u_k}(0,s) =2\int_{B(0,t)\setminus B(0,s)} |y|^{2-m}\left|\frac{\partial u_k}{\partial r}(y)\right|^2\,d\mathcal L^m(y), \end{align*} where $\partial u_k/\partial r$ is the [weak derivative](/page/Weak%20Derivative) in the radial direction $y/|y|$. The right-hand side measures exactly how far $u_k$ is from being constant along rays. The scaling is compatible with the normalized energy: for every fixed $\rho>0$ and all sufficiently large $k$, \begin{align*} \Theta_{u_k}(0,\rho)=\Theta_u(a,r_k\rho). \end{align*} Since $r_k\rho\downarrow0$, both $\Theta_u(a,r_kt)$ and $\Theta_u(a,r_ks)$ converge to the same density $\Theta_u(a,0^+)$. Hence \begin{align*} \lim_{k\to\infty}\left(\Theta_{u_k}(0,t)-\Theta_{u_k}(0,s)\right)=0. \end{align*} Because $u_k\to\nu_0$ strongly in $W^{1,2}_{\mathrm{loc}}(\mathbb R^m;\mathbb R^q)$, the gradients converge strongly in $L^2$ on the fixed annulus $A_{s,t}:=B(0,t)\setminus B(0,s)$. The radial derivative is obtained by projecting the gradient onto the unit vector $y/|y|$, and this projection has operator norm at most $1$. Hence $h_k:=\partial u_k/\partial r$ converges strongly to $h:=\partial\nu_0/\partial r$ in $L^2(A_{s,t};\mathbb R^q)$. We still need to justify convergence of the squared radial energies. Use the pointwise inequality \begin{align*} \big||h_k|^2-|h|^2\big|\le |h_k-h|(|h_k|+|h|). \end{align*} After integrating and applying the Cauchy-Schwarz inequality, \begin{align*} \int_{A_{s,t}}\big||h_k(y)|^2-|h(y)|^2\big|\,d\mathcal L^m(y) \le \|h_k-h\|_{L^2(A_{s,t})}\bigl(\|h_k\|_{L^2(A_{s,t})}+\|h\|_{L^2(A_{s,t})}\bigr). \end{align*} The first factor tends to $0$, and the second factor stays bounded because strong convergence implies boundedness. Therefore $|h_k|^2\to |h|^2$ in $L^1(A_{s,t})$. Since the annulus stays away from the origin, the weight $y\mapsto |y|^{2-m}$ is bounded on $A_{s,t}$, so the weighted squared radial energies also converge. Passing to the limit in the monotonicity identity gives \begin{align*} 2\int_{B(0,t)\setminus B(0,s)} |y|^{2-m}\left|\frac{\partial \nu_0}{\partial r}(y)\right|^2\,d\mathcal L^m(y)=0. \end{align*} Since the integrand is nonnegative, $\partial \nu_0/\partial r=0$ for $\mathcal L^m$-a.e. point of the annulus. The annulus was arbitrary, so this holds on $\mathbb R^m\setminus\{0\}$. Therefore $\nu_0$ is constant along almost every ray, which is precisely the homogeneity relation \begin{align*} \nu_0(\lambda y)=\nu_0(y) \end{align*} for every $\lambda>0$ and for $\mathcal L^m$-a.e. $y\in\mathbb R^m$. Finally, the tangent map cannot be constant. Suppose that $\nu_0$ were constant. Because $u_k\to\nu_0$ strongly in $W^{1,2}(B(0,2);\mathbb R^q)$, the gradients converge strongly in $L^2(B(0,2);\mathbb R^{q\times m})$, and therefore \begin{align*} \lim_{k\to\infty}\int_{B(0,2)} |\nabla u_k(y)|^2\,d\mathcal L^m(y)=\int_{B(0,2)} |\nabla \nu_0(y)|^2\,d\mathcal L^m(y)=0. \end{align*} The normalized energies are related by scaling: \begin{align*} \Theta_u(a,2r_k)=\Theta_{u_k}(0,2)=2^{2-m}\int_{B(0,2)} |\nabla u_k(y)|^2\,d\mathcal L^m(y). \end{align*} The right-hand side tends to $0$, while $2r_k\downarrow0$, so the left-hand side tends to the density $\Theta_u(a,0^+)$. Hence $\Theta_u(a,0^+)=0$, contradicting the singular-point lower bound $\Theta_u(a,0^+)\ge\varepsilon_0$. Thus the tangent map is a nonconstant homogeneous minimizing map.[/guided]

[step:Apply dimension reduction to rule out strata above dimension $m-3$]For each integer $j\in\{0,1,\dots,m\}$, define the $j$-th symmetry stratum $S_j(u)\subset\operatorname{sing}(u)$ as follows: a point $a\in\operatorname{sing}(u)$ belongs to $S_j(u)$ if every tangent map at $a$, viewed as a map on the tangent domain $\mathbb R^m$, fails to be invariant under translations by every linear subspace $V\subset\mathbb R^m$ with $\dim V=j+1$. Here invariance under $V$ means $\nu(y+v)=\nu(y)$ for every $v\in V$ and for $\mathcal L^m$-a.e. $y\in\mathbb R^m$. The [Schoen-Uhlenbeck dimension reduction theorem](/page/Schoen-Uhlenbeck%20Dimension%20Reduction%20Theorem) states that \begin{align*} \dim_{\mathcal H} S_j(u)\le j. \end{align*} It remains to prove that $\operatorname{sing}(u)\subset S_{m-3}(u)$ when $m\ge3$. Assume, toward contradiction, that $a\in\operatorname{sing}(u)$ has a tangent map $\nu_0\in W^{1,2}_{\mathrm{loc}}(\mathbb R^m;N)$ invariant under an $(m-2)$-dimensional linear subspace $V\subset\mathbb R^m$. Let $\mathcal L^2_{V^\perp}$ denote two-dimensional Lebesgue measure on the Euclidean plane $V^\perp$, and let $P:V^\perp\to\mathbb R^m$ denote the inclusion of the orthogonal complement of $V$. For $r>0$, let $B_V(0,r)$ and $B_{V^\perp}(0,r)$ denote the open Euclidean balls of radius $r$ centered at $0$ in the subspaces $V$ and $V^\perp$, respectively. Define $w:V^\perp\setminus\{0\}\to N$ by $w(z):=\nu_0(Pz)$. The [translation invariance](/theorems/4911) of $\nu_0$ along $V$ implies that $\nu_0(y)=w(\pi_{V^\perp}y)$ for $\mathcal L^m$-a.e. $y\in\mathbb R^m$, where $\pi_{V^\perp}:\mathbb R^m\to V^\perp$ is the [orthogonal projection](/theorems/437). We record two consequences of this representation. First, $w$ is a Sobolev map with finite energy near the puncture. Choose coordinates $(v,z)\in V\times V^\perp$ and write $F:V\times V^\perp\to\mathbb R^m$ for the linear isometry $F(v,z)=v+z$. Since $\nu_0\circ F=w\circ\pi_{V^\perp}$ almost everywhere and $\nu_0\in W^{1,2}_{\mathrm{loc}}(\mathbb R^m;\mathbb R^q)$, the weak chain rule for linear maps gives, after redefining $w$ on a null set, $w\in W^{1,2}_{\mathrm{loc}}(V^\perp\setminus\{0\};N)$ and \begin{align*} \nabla(\nu_0\circ F)(v,z)=\nabla w(z)\circ \pi_{V^\perp} \end{align*} for $\mathcal L^{m-2}\otimes\mathcal L^2_{V^\perp}$-a.e. $(v,z)$ away from $z=0$. The integrand $|\nabla\nu_0|^2$ is integrable on compact cylinders because $\nu_0\in W^{1,2}_{\mathrm{loc}}$. Therefore Tonelli's theorem and the preceding gradient identity give, for $0<\rho<1$ and $C_\rho:=\{v+z:v\in V, |v|<1, z\in B_{V^\perp}(0,\rho)\}$, \begin{align*} \mathcal L^{m-2}(B_V(0,1))\int_{B_{V^\perp}(0,\rho)} |\nabla w(z)|^2\,d\mathcal L^2_{V^\perp}(z) = \int_{C_\rho} |\nabla \nu_0(y)|^2\,d\mathcal L^m(y)<\infty. \end{align*} Choose an oriented [orthonormal basis](/page/Orthonormal%20Basis) $(e_1,e_2)$ of $V^\perp$ and define the polar parametrization $\Phi:(0,\infty)\times(0,2\pi)\to V^\perp\setminus\{0\}$ by \begin{align*} \Phi(r,\theta):=r(\cos\theta\,e_1+\sin\theta\,e_2). \end{align*} The Sobolev chain rule gives $W:=w\circ\Phi\in W^{1,2}_{\mathrm{loc}}((0,\infty)\times(0,2\pi);N)$. By Fubini, choose the polar representative of $W$ for which the a.e. homogeneity relation becomes $W(r,\theta)=W(1,\theta)$ for $\mathcal L^1\otimes\mathcal L^1$-a.e. $(r,\theta)$. Define $\partial_\theta w$ to be the weak $\theta$-derivative of the trace $\theta\mapsto W(1,\theta)$ on $\partial B_{V^\perp}(0,1)$. The weak chain rule in polar coordinates then gives, for $\mathcal L^1\otimes\mathcal L^1$-a.e. $(r,\theta)$, \begin{align*} |\nabla w(\Phi(r,\theta))|^2=r^{-2}|\partial_\theta w(\theta)|^2. \end{align*} The polar-coordinate formula on $V^\perp$ uses the substitution $z=r(\cos\theta\,e_1+\sin\theta\,e_2)$, under which $d\mathcal L^2_{V^\perp}(z)=r\,d\mathcal L^1(r)\,d\mathcal L^1(\theta)$. Hence, for every $0<\rho<R<\infty$, \begin{align*} \int_{B_{V^\perp}(0,R)\setminus B_{V^\perp}(0,\rho)} |\nabla w(z)|^2\,d\mathcal L^2_{V^\perp}(z) =\log(R/\rho)\int_{\partial B_{V^\perp}(0,1)} |\partial_\theta w|^2\,d\mathcal H^1. \end{align*} The finite-energy estimate above allows $\rho\downarrow0$ with $R$ fixed. If the boundary integral were positive, the right-hand side would diverge like $\log(R/\rho)$, contradicting finite energy on $B_{V^\perp}(0,R)$. Therefore $\partial_\theta w=0$ for $\mathcal H^1$-a.e. point of $\partial B_{V^\perp}(0,1)$. Since $w$ is also homogeneous, it is constant for $\mathcal L^2_{V^\perp}$-a.e. point of $V^\perp\setminus\{0\}$. Hence $\nu_0$ is constant, contradicting the nonconstancy of tangent maps at singular points proved above. Therefore no tangent map at any singular point has an $(m-2)$-dimensional translation symmetry. This means $\operatorname{sing}(u)\subset S_{m-3}(u)$ for $m\ge3$. Applying dimension reduction with $j=m-3$ gives \begin{align*} \dim_{\mathcal H}\operatorname{sing}(u)\le m-3. \end{align*}[/step]

[guided]Assume $m\ge3$ for this dimension-reduction step. The dimension estimate comes from excluding too much symmetry in tangent maps. For an integer $j\in\{0,1,\dots,m\}$, define $S_j(u)$ to be the set of singular points $a$ such that every tangent map at $a$ fails to be invariant under translations by every linear subspace $V\subset\mathbb R^m$ with $\dim V=j+1$. The word linear matters: tangent maps live on the [vector space](/page/Vector%20Space) $\mathbb R^m$ based at the blow-up point, so the symmetry directions pass through the origin. The [Schoen-Uhlenbeck dimension reduction theorem](/page/Schoen-Uhlenbeck%20Dimension%20Reduction%20Theorem) gives \begin{align*} \dim_{\mathcal H}S_j(u)\le j. \end{align*} Thus it suffices to show that every singular point belongs to $S_{m-3}(u)$. Suppose instead that some singular point $a$ has a tangent map $\nu_0$ invariant under an $(m-2)$-dimensional linear subspace $V\subset\mathbb R^m$. The quotient directions form the two-dimensional orthogonal complement $V^\perp$. Let $\mathcal L^2_{V^\perp}$ denote the Lebesgue measure on this Euclidean plane. Let $P:V^\perp\to\mathbb R^m$ be the inclusion map. For $r>0$, write $B_V(0,r)$ and $B_{V^\perp}(0,r)$ for the open Euclidean balls of radius $r$ centered at $0$ in $V$ and $V^\perp$, respectively. Define $w:V^\perp\setminus\{0\}\to N$ by $w(z):=\nu_0(Pz)$. Because $\nu_0$ is invariant under translations along $V$, all of its variation occurs in the two transverse variables. Equivalently, if $\pi_{V^\perp}:\mathbb R^m\to V^\perp$ is the orthogonal projection, then $\nu_0(y)=w(\pi_{V^\perp}y)$ for $\mathcal L^m$-a.e. $y$. Before applying any two-dimensional argument, we need finite energy near the puncture. This does not follow merely from finite energy on annuli. It follows from the fact that $\nu_0\in W^{1,2}_{\mathrm{loc}}(\mathbb R^m;N)$ and is independent of the $V$-variables. For $0<\rho<1$, define the cylinder $C_\rho:=\{v+z:v\in V, |v|<1, z\in B_{V^\perp}(0,\rho)\}$. [Fubini's theorem](/theorems/2961) gives \begin{align*} \mathcal L^{m-2}(B_V(0,1))\int_{B_{V^\perp}(0,\rho)} |\nabla w(z)|^2\,d\mathcal L^2_{V^\perp}(z) \le \int_{C_\rho} |\nabla\nu_0(y)|^2\,d\mathcal L^m(y)<\infty. \end{align*} Thus $w$ has finite Dirichlet energy on punctured disks together with the puncture included in the energy integral. The homogeneity of $\nu_0$ descends to $w$: for every $\lambda>0$, \begin{align*} w(\lambda z)=w(z) \end{align*} for $\mathcal L^2_{V^\perp}$-a.e. $z\in V^\perp$. Choose an oriented orthonormal basis $(e_1,e_2)$ of $V^\perp$ and define the polar parametrization $\Phi:(0,\infty)\times(0,2\pi)\to V^\perp\setminus\{0\}$ by \begin{align*} \Phi(r,\theta):=r(\cos\theta\,e_1+\sin\theta\,e_2). \end{align*} The Sobolev chain rule gives $W:=w\circ\Phi\in W^{1,2}_{\mathrm{loc}}((0,\infty)\times(0,2\pi);N)$. Since the homogeneity statement holds only almost everywhere in $z$, we choose a polar representative using Fubini: after changing $W$ on an $\mathcal L^1\otimes\mathcal L^1$-null set, $W(r,\theta)=W(1,\theta)$ for a.e. $(r,\theta)$. Define $\partial_\theta w$ as the weak derivative of the trace $\theta\mapsto W(1,\theta)$. The weak chain rule in polar coordinates then gives, for a.e. $(r,\theta)$, \begin{align*} |\nabla w(\Phi(r,\theta))|^2=r^{-2}|\partial_\theta w(\theta)|^2. \end{align*} In polar coordinates on $V^\perp$, the substitution $z=r(\cos\theta\,e_1+\sin\theta\,e_2)$ transforms measure by $d\mathcal L^2_{V^\perp}(z)=r\,d\mathcal L^1(r)\,d\mathcal L^1(\theta)$. Thus, for $0<\rho<R<\infty$, \begin{align*} \int_{B_{V^\perp}(0,R)\setminus B_{V^\perp}(0,\rho)} |\nabla w(z)|^2\,d\mathcal L^2_{V^\perp}(z) =\log(R/\rho)\int_{\partial B_{V^\perp}(0,1)} |\partial_\theta w|^2\,d\mathcal H^1. \end{align*} Now use the finite-energy estimate as $\rho\downarrow0$. If the angular energy on $\partial B_{V^\perp}(0,1)$ were positive, the right-hand side would diverge logarithmically. This contradicts finite energy of $w$ on $B_{V^\perp}(0,R)$. Therefore $\partial_\theta w=0$ for $\mathcal H^1$-a.e. point of the unit circle. Combined with homogeneity, this implies that $w$ is constant for $\mathcal L^2_{V^\perp}$-a.e. point of $V^\perp\setminus\{0\}$. Therefore $\nu_0$ is constant, contradicting the nonconstant tangent map obtained at a singular point. This contradiction proves, for $m\ge3$, that no singular point admits a tangent map with $(m-2)$ independent translation symmetries. Therefore $\operatorname{sing}(u)\subset S_{m-3}(u)$, and dimension reduction gives \begin{align*} \dim_{\mathcal H}\operatorname{sing}(u)\le m-3. \end{align*}[/guided]

[step:Derive the Hausdorff measure conclusion]If $m\ge3$, the estimate $\dim_{\mathcal H}\operatorname{sing}(u)\le m-3$ implies \begin{align*} \mathcal H^{m-2}(\operatorname{sing}(u))=0, \end{align*} because [Hausdorff measure](/page/Hausdorff%20Measure) in every dimension strictly larger than the Hausdorff dimension vanishes. If $m=2$, the [two-dimensional regularity theorem for minimizing harmonic maps](/page/Two-Dimensional%20Regularity%20Theorem%20for%20Minimizing%20Harmonic%20Maps) applies to the local minimizing map $u$ on planar balls compactly contained in $U$ and gives $\operatorname{sing}(u)=\varnothing$. Therefore \begin{align*} \mathcal H^0(\operatorname{sing}(u))=0. \end{align*} Combining the previous steps proves smoothness on $U\setminus\operatorname{sing}(u)$, relative closedness of $\operatorname{sing}(u)$, the estimate $\mathcal H^{m-2}(\operatorname{sing}(u))=0$, and, for $m\ge3$, the sharper bound $\dim_{\mathcal H}\operatorname{sing}(u)\le m-3$.[/step]

[guided]There are two cases because the dimension-reduction step gives the sharp estimate only when $m\ge3$. In that case we already proved \begin{align*} \dim_{\mathcal H}\operatorname{sing}(u)\le m-3. \end{align*} Hausdorff measure in dimension $s$ vanishes on every set whose Hausdorff dimension is strictly smaller than $s$. Taking $s=m-2$ gives \begin{align*} \mathcal H^{m-2}(\operatorname{sing}(u))=0. \end{align*} When $m=2$, the theorem statement asks for $\mathcal H^0(\operatorname{sing}(u))=0$. The [two-dimensional regularity theorem for minimizing harmonic maps](/page/Two-Dimensional%20Regularity%20Theorem%20for%20Minimizing%20Harmonic%20Maps) applies on every ball $B\Subset U$, since $u$ is a local energy minimizer and $N$ is compact. It gives smoothness on each such ball, so $\operatorname{sing}(u)=\varnothing$. Hence \begin{align*} \mathcal H^0(\operatorname{sing}(u))=0. \end{align*} The regular set was defined using epsilon regularity, so $u$ is smooth on $U\setminus\operatorname{sing}(u)$. The openness of the regular set proved above makes $\operatorname{sing}(u)$ relatively closed in $U$. These conclusions are exactly the asserted partial regularity, Hausdorff-measure vanishing, and, for $m\ge3$, the sharper Hausdorff-dimension estimate.[/guided]