Subellipticity and unique continuation · Self-checked by the writing AI

Quadratic energy and the positive trace

The harmonic oscillator has positive energy even though its classical energy vanishes at the origin. For a general nonnegative quadratic form, the same correction is the sum of its positive symplectic frequencies. Directions that contribute only a multiplication square have a different effect: they can prevent a minimum from being attained without raising the infimum.

We use the positive definite symplectic normalization in Section 7 of When a nonnegative scalar symbol acquires a negative part, and the unitary covariance of Weyl quantization in Section 6 of From Weyl symbols to operators and changes of coordinates. The extension to a form with a kernel, its exact operator identity, and its sharp lower bound are proved here. Nicacio [N] gives an open introduction to positive definite symplectic normalization; Kamat and Mishra [K] discuss its relation to more general quadratic forms. The correction below is the quadratic ingredient in Melin's lower bounds [M].

1. Conventions and the Hamilton map

Put \(z=(x,\xi)\in\mathbb R^{2n}\), and use \[ \omega((x,\xi),(y,\eta))=\xi\cdot y-x\cdot\eta, \qquad D=-i\partial_x. \tag{1.1} \] For functions on phase space our bracket is \[ \{a,b\}=\partial_\xi a\cdot\partial_x b -\partial_x a\cdot\partial_\xi b; \qquad \{x_j,\xi_k\}=-\delta_{jk}. \tag{1.2} \] The first Weyl product correction is \((2i)^{-1}\{a,b\}\). In particular \([D_j,x_k]=-i\delta_{jk}\).

Let \(Q(z)=z^THz\), where \(H\) is real symmetric and nonnegative. Its polar form is \(Q(u,v)=u^THv\); thus no factor of two enters the polar form. Define the Hamilton map by \[ \omega(u,Fv)=Q(u,v). \tag{1.3} \] In these coordinates, with \[ J=\begin{pmatrix}0&-I\\ I&0\end{pmatrix}, \qquad K=-J, \] we have \(F=KH\). Two identities will be useful: \[ \omega(Fu,v)+\omega(u,Fv)=0, \qquad \ker F=\ker H=\{v:Q(v)=0\}. \tag{1.4} \] The first follows because \(JF=H\) is symmetric. The second follows from invertibility of \(K\) and the spectral decomposition of a nonnegative symmetric matrix.

The nonzero eigenvalues of \(F\) will be \(\pm i\mu_1,\ldots,\pm i\mu_k\), with \(\mu_j>0\), including multiplicities. We define \[ \operatorname{Tr}_+Q=\sum_{j=1}^k\mu_j. \tag{1.5} \] This is a definition through the Hamilton map, rather than through the ordinary eigenvalues of \(H\).

2. What happens when the quadratic form has a kernel

Theorem 2.1 (semidefinite symplectic normal form). There are symplectic linear coordinates and integers \(k,\ell\geq0\), \(k+\ell\leq n\), in which \[ Q(x,\xi)=\sum_{j=1}^k\mu_j(x_j^2+\xi_j^2) +\sum_{j=k+1}^{k+\ell}x_j^2. \tag{2.1} \] The remaining coordinate pairs do not occur. The nonzero eigenvalues of \(F\) are the pairs in (1.5), and are semisimple. On its generalized zero eigenspace, \(F^2=0\).

Proof. Let \(T=H^{1/2}\) and \(B=TKT\). The real matrix \(B\) is skew-adjoint. Consequently its complexification is diagonalizable, with eigenvalues zero and pairs \(\pm i\mu\), \(\mu>0\). We have \[ TF=BT. \tag{2.2} \] The matrices \(F=(KT)T\) and \(B=T(KT)\) have the same characteristic polynomial. To see this directly, block elimination in \(\begin{pmatrix}I&A\\ B_0&I\end{pmatrix}\) gives \(\det(I-AB_0)=\det(I-B_0A)\); apply this with \(A=KT/\lambda\), \(B_0=T\), and then use polynomial equality in \(\lambda\).

On any generalized eigenspace of \(F\) for \(\lambda\ne0\), \(T\) is injective: \(Tv=0\) implies \(Fv=0\), and \(F\) is invertible on that space. Equation (2.2) maps it to the corresponding eigenspace of \(B\). Hence those generalized eigenspaces have no nontrivial Jordan blocks. Let \(U\) be their real direct sum. The form \(Q\) is positive definite on \(U\), since \(T\) is injective there.

The subspace \(U\) is symplectic. Indeed, if \(u\in U\) is symplectically orthogonal to \(U\), then \(FU=U\) and (1.3) give \(Q(u,U)=0\), so \(u=0\). Write \(W=U^{\perp_\omega}\). Identity (1.4) makes \(W\) invariant under \(F\), and (1.3) also makes it \(Q\)-orthogonal to \(U\). All nonzero eigenspaces have already been included in \(U\). Thus \(F|_W\) is nilpotent. Equation (2.2) maps \(TW\) into the generalized zero eigenspace of the diagonalizable matrix \(B\), which is \(\ker B\). Therefore \[ TFw=BTw=0,\qquad F^2w=0\quad(w\in W). \tag{2.3} \]

Apply the positive definite symplectic normalization from the prerequisite to \(Q|_U\). It gives the oscillator part of (2.1); computing its Hamilton map identifies its coefficients as \(\mu_j\).

It remains to treat \(W\). Put \(N=\ker F\subset W\). The induced form on \(W/N\) is positive definite. Let \(v_1,\ldots,v_\ell\) represent an orthonormal basis for it, and put \(w_j=Fv_j\). Equations (1.3), (1.4), and (2.3) give \[ \omega(v_a,w_b)=\delta_{ab},\qquad \omega(w_a,w_b)=0,\qquad Q(w_a,\cdot)=0. \tag{2.4} \] Write \(A_{ab}=\omega(v_a,v_b)\). Replace \[ v_a\quad\text{by}\quad v'_a=v_a+\frac12\sum_b A_{ab}w_b. \tag{2.5} \] Since \(A\) is skew-symmetric, direct substitution gives \(\omega(v'_a,v'_b)=A_{ab}+A_{ba}/2-A_{ab}/2=0\). The other identities in (2.4), and all \(Q\)-products, are unchanged. Thus the pairs \((v'_j,-w_j)\) have symplectic products exactly as the coordinate pairs \((e_j,\varepsilon_j)\) in (1.1), and \(Q\) is the sum of their \(x\)-squares.

Their span is symplectic. On its symplectic orthogonal complement in \(W\), \(Q\) vanishes: a vector in that complement is \(Q\)-orthogonal to every \(v'_j\) by (1.3), and these vectors span \(W/N\). Complete the complement to a symplectic basis by successively choosing a nonzero vector and a vector with nonzero symplectic product, normalizing that product, and passing to the orthogonal complement of the resulting plane. This constructs the free coordinate pairs. It also proves \(k+\ell\leq n\). ∎

The example \(Q=x^2\) has a nonzero Hessian but only zero Hamilton eigenvalues. Its Hamilton map is a nonzero nilpotent map. This is why a positive definite normal form alone does not cover semidefinite forms.

3. An invariant sum of operator squares

Let \(V_+\) be the sum of the complex eigenspaces of \(F\) with eigenvalues \(i\mu\), \(\mu>0\), and let \(V_0\) be its generalized zero eigenspace. The preceding proof shows that \[ \frac12Q(\overline v,v) \tag{3.1} \] is a positive Hermitian form on \(V_+\). Choose a unitary basis \(v_1,\ldots,v_k\) for it. Choose representatives \(v_{k+1},\ldots,v_{k+\ell}\) for a real \(Q\)-orthonormal basis of \(V_0/\ker F\). Define the complex linear functions \[ L_j(z)=Q(v_j,z). \tag{3.2} \] Adding a vector in \(\ker F\) to a real representative does not change \(L_j\).

Theorem 3.1 (exact quadratic identity). These functions satisfy \[ \sum_j|L_j(z)|^2=Q(z),\qquad \sum_j\{\operatorname{Re}L_j,\operatorname{Im}L_j\} =-\operatorname{Tr}_+Q, \tag{3.3} \] and, on Schwartz functions, \[ Q^w=\sum_j(L_j^w)^*L_j^w+\operatorname{Tr}_+Q. \tag{3.4} \]

Proof. In (2.1) choose \(v_j=\mu_j^{-1/2}(e_j+i\varepsilon_j)\) for \(j\leq k\), and \(v_j=e_j\) in the multiplication directions. Then \[ L_j=\sqrt{\mu_j}(x_j+i\xi_j)\quad(j\leq k), \qquad L_j=x_j\quad(k<j\leq k+\ell). \tag{3.5} \] Both identities (3.3) follow, using \(\{x_j,\xi_j\}=-1\).

Any other unitary basis of \(V_+\) changes this list by a unitary matrix. Its sum of absolute squares and its sum of operator squares are unchanged. A different orthonormal basis of the real quotient changes the remaining list by an orthogonal matrix. Symplectic coordinate changes preserve the bracket, so (3.3) holds in the original coordinates as well.

For a linear function \(L=a+ib\), with \(a,b\) real, the Weyl product terminates after its first correction: \[ \overline L\mathbin\#L =a^2+b^2+\{a,b\}. \tag{3.6} \] Indeed \(\{a-ib,a+ib\}=2i\{a,b\}\). Weyl adjunction is complex conjugation, so summing (3.6) and using (3.3) proves (3.4). ∎

For a real quadratic polynomial, the symmetric part of its left quantization equals its Weyl quantization. The conversion correction is purely imaginary and constant; taking the symmetric part removes it. Thus (3.4) also describes the symmetric part if one starts with a left symbol.

4. The lower bound is sharp

Theorem 4.1. For every nonzero Schwartz function, \[ \frac{(Q^wu,u)}{\|u\|^2}\geq\operatorname{Tr}_+Q, \qquad \inf_{0\ne u\in\mathcal S} \frac{(Q^wu,u)}{\|u\|^2}=\operatorname{Tr}_+Q. \tag{4.1} \] If \(\ell>0\), the infimum is not attained by a nonzero \(L^2\) function in the quadratic form domain. If \(\ell=0\), it is attained. In the normal coordinates its minimizing space consists of the oscillator ground state in the first \(k\) variables tensored with arbitrary \(L^2\) functions in the free variables.

Proof. The lower bound follows immediately from (3.4). In the normal coordinates let \[ g(t)=\pi^{-1/4}e^{-t^2/2},\qquad g_\varepsilon(t)=\varepsilon^{-1/2}g(t/\varepsilon). \tag{4.2} \] We have \(\|g\|=\|g_\varepsilon\|=1\), \((t+\partial_t)g=0\), and \(\|t g_\varepsilon\|^2=\varepsilon^2/2\). Take the product of \(g\) in the oscillator variables, \(g_\varepsilon\) in the multiplication variables, and any normalized Schwartz function in the free variables. Its quotient in (4.1) is \[ \sum_{j=1}^k\mu_j+\frac\ell2\varepsilon^2. \tag{4.3} \] The unitary symplectic covariance from the prerequisite transports these test functions to the original coordinates and preserves Schwartz space. Letting \(\varepsilon\downarrow0\) proves sharpness.

In normal coordinates the sum-of-squares form is closed on the intersection of the domains of \(\partial_{x_j}\), \(x_j\) for oscillator variables, and \(x_j\) for multiplication variables. Schwartz functions are a form core: smooth compact truncation followed by convolution converges in all these graph norms. In detail, truncation errors for derivatives contain a bounded derivative of the cutoff tending to zero, while multiplication errors are controlled by the corresponding weighted \(L^2\) tails. Convolution converges in derivative norms, and \(x_j(u*\rho_\varepsilon)=(x_ju)*\rho_\varepsilon+u*(x_j\rho_\varepsilon)\), whose last term tends to zero in \(L^2\). These facts also identify the closure after unitary transport.

Equality in (3.4) forces every \(L_j^wu\) to vanish. If a multiplication direction is present, \(x_j u=0\) almost everywhere forces \(u=0\), because its zero hyperplane has Lebesgue measure zero. If none is present, the oscillator equations are \((x_j+\partial_{x_j})u=0\). Multiplication by \(e^{|x'|^2/2}\) makes the weak derivatives in those variables zero, so their solutions are \(g^{\otimes k}\otimes f\), with \(f\in L^2\) in the free variables. This proves the attainment assertions. ∎

The closed form defines a nonnegative self-adjoint operator by the usual quadratic form construction; (4.1) is also its spectral infimum. No description of its entire spectrum is needed for (4.1).

5. Continuity, scaling, and smooth families

There is a matrix formula that remains meaningful when ranks change: \[ \operatorname{Tr}_+Q =\frac12\operatorname{tr}\sqrt{-B^2}, \qquad B=H^{1/2}KH^{1/2}. \tag{5.1} \] The eigenvalues of \(\sqrt{-B^2}\) are two copies of each \(\mu_j\) and zeros, so (5.1) follows from the proof of Theorem 2.1. The square root of a nonnegative matrix is continuous: on any bounded spectral interval approximate \(t\mapsto\sqrt t\) uniformly by polynomials, apply those polynomials to the matrices, and use the finite-dimensional spectral theorem to bound the approximation error. Applying this twice in (5.1) proves continuity of \(\operatorname{Tr}_+Q\), even at changes of rank. Also \[ \operatorname{Tr}_+(aQ)=a\operatorname{Tr}_+Q\quad(a\geq0). \tag{5.2} \] A symplectic change of basis conjugates \(F\), proving symplectic invariance.

For later use suppose \(Q_\rho\) is a smooth family with constant rank, and the symplectic form restricted to \(N_\rho=\ker Q_\rho\) has constant rank. In (2.1), \[ \dim N_\rho=2n-2k-\ell, \qquad \operatorname{rank}(\omega|_{N_\rho})=2(n-k-\ell). \tag{5.3} \] Thus \(k\) and \(\ell\) are constant. The spaces \(V_+\), \(V_0\), and \(N\) form smooth local bundles. For completeness, surround the positive imaginary eigenvalues at one parameter by contours separated from zero and the negative eigenvalues. The matrix \((2\pi i)^{-1}\int (\zeta-F_\rho)^{-1}\,d\zeta\) is a smooth projection for nearby parameters; evaluation in a Jordan basis shows that its image is exactly \(V_+\). A contour around zero gives \(V_0\). Constant rank gives smooth local bases for \(N\) by an invertible minor. Project fixed bases with these projections and use Gram–Schmidt for the positive Hermitian form (3.1), or for the real positive quotient form. This constructs smooth local choices of the \(L_j\) in (3.2). Coinciding nonzero frequencies cause no difficulty, because the projection surrounds their whole positive group.

Finally let \(p_r\geq0\) be homogeneous of degree \(r\) in \(\xi\), and vanish at \(\rho\ne0\). Then \(dp_r(\rho)=0\), and \[ Q_\rho(v)=\frac12d^2p_r(\rho)[v,v] \tag{5.4} \] is intrinsic. For \(D_\lambda(x,\xi)=(x,\lambda\xi)\), \[ Q_{D_\lambda\rho}(D_\lambda v)=\lambda^rQ_\rho(v), \qquad D_\lambda^*\omega=\lambda\omega. \] Equation (1.3) therefore gives \[ F_{D_\lambda\rho}D_\lambda =\lambda^{r-1}D_\lambda F_\rho, \qquad \operatorname{Tr}_+Q_{D_\lambda\rho} =\lambda^{r-1}\operatorname{Tr}_+Q_\rho. \tag{5.5} \] The correction has the degree of a subprincipal symbol.

6. Exercises and complete solutions

Exercise 1 — a mixed quadratic form, 8 points. Let \(Q(x,\xi)=a x^2+2b x\xi+c\xi^2\), with \(a,c>0\) and \(ac>b^2\). Compute \(\operatorname{Tr}_+Q\) and a minimizing function.

Solution. Here \(F=\begin{pmatrix}b&c\\-a&-b\end{pmatrix}\), so \(F^2=-(ac-b^2)I\). Hence \(\operatorname{Tr}_+Q=\mu=\sqrt{ac-b^2}\). Complete the square: \[ Q^w=c(D+(b/c)x)^2+(a-b^2/c)x^2. \] For \(u=e^{-ibx^2/(2c)}v\), the first factor becomes \(e^{-ibx^2/(2c)}Dv\). Choose \(v=C e^{-\mu x^2/(2c)}\), with \(C\) normalizing its \(L^2\) norm. Direct differentiation then gives \(Q^wu=\mu u\), as required.

Exercise 2 — disappearance of a frequency, 8 points. For \(t\geq0\), take \(Q_t=x^2+t^2\xi^2\). Find its positive trace, and explain the change in attainment at \(t=0\).

Solution. For \(t>0\), \(F_t^2=-t^2I\), so \(\operatorname{Tr}_+Q_t=t\), with normalized ground state proportional to \(e^{-x^2/(2t)}\). At \(t=0\), \(Q_0^w=x^2\); its positive trace and infimum are zero. Normalized functions concentrated near zero approach that infimum, but \(x u=0\) has no nonzero \(L^2\) solution. The trace is continuous across the rank change, although a normalized ground state has no \(L^2\) limit.

Exercise 3 — the Planck parameter, 6 points. Quantize with \(hD\), \(h>0\). State and prove the analogue of (4.1) for a fixed nonnegative quadratic form.

Solution. The first Weyl product correction becomes \(h(2i)^{-1}\{a,b\}\), so (3.4) becomes \(Q_h^w=\sum(L_{j,h}^w)^*L_{j,h}^w+h\operatorname{Tr}_+Q\). The lower bound is \(h\operatorname{Tr}_+Q\). In normal coordinates use the Gaussian of width \(\sqrt h\) in oscillator directions and widths tending to zero in multiplication directions. Their quotients tend to the same bound. Symplectic covariance holds with this fixed quantization parameter as well.

Exercise 4 — a radical that is not symplectic, 10 points. For the normal form (2.1), compute the radical of \(\omega|_{\ker Q}\). Prove that \(\ker Q\) is symplectic exactly when \(\ell=0\), and relate this to diagonalizability of \(F\).

Solution. The kernel consists of the \(\xi_j\)-axes for multiplication variables and the entire free coordinate planes. The free planes are symplectic and mutually orthogonal. The \(\xi_j\)-axes pair to zero with the entire kernel, so they span its symplectic radical, of dimension \(\ell\). Thus the kernel is symplectic exactly when \(\ell=0\). Each multiplication plane has \(F e_j=-\varepsilon_j\), \(F\varepsilon_j=0\), a nontrivial zero Jordan block. The oscillator planes are diagonalizable over \(\mathbb C\), and the free planes have zero map. Consequently \(F\) is diagonalizable over \(\mathbb C\) exactly when \(\ell=0\).

References

Written by GPT-6.1 Sol (OpenAI), at Ultra reasoning effort, September 2026. Self-checked by the writing AI. Public domain (CC0).