Metric operations

Building a metric that is suitable for remeshing may require combining information from different sources or imposing constraints (e.g. minimum or maximum edge lengths, smooth variation with respect to the input mesh, ... ),

Metric intersection

Given two metrics ${\mathcal{M}}_1$ and ${\mathcal{M}}_2$, the intersection ${\mathcal{M}}_{1\cap 2}$ corresponds to a metric that imposes the largest sizes that are smaller than those imposed by ${\mathcal{M}}_1$ and ${\mathcal{M}}_2$: $${\mathcal{M}}_{1\cap 2}=\arg \min \{\det (\mathcal{M})|\mathcal{M}\in {\mathcal{S}}^{+}s.t.\mathcal{M}\ge {\mathcal{M}}_1,\mathcal{M}\ge {\mathcal{M}}_2\}$$

It is computed as follows: let $\mathcal{P}=(e_0|...|e_d)$ be the generalized eigenvectors of $({\mathcal{M}}_0,{\mathcal{M}}_1)$: $${\mathcal{M}}_0\mathcal{P}=\Lambda {\mathcal{M}}_1\mathcal{P}$$ then $${\mathcal{M}}_i={\mathcal{P}}^{-1,T}{\Lambda }^{(i)}{\mathcal{P}}^{(-1)}$$ with ${\Lambda }_{jk}^{(i)}=e_j^T{\mathcal{M}}_i{e}_k{\delta }_{jk}$. The intersection is then $${\mathcal{M}}_0\cap {\mathcal{M}}_1={\mathcal{P}}^{-1,T}{\Lambda }^{(i,j)}{\mathcal{P}}^{-1}$$ with ${\Lambda }_{jk}^{(i,j)}=\max ({\Lambda }_{jk}^{(i)},{\Lambda }_{jk}^{(j)})$.

See proof

Metric gradation

It is sometimes desirable to control how fast edge lengths grow from from element to the next. This idea can be translated into constraints on the variation of the metric field on the edges of the mesh (see Size gradation control of anisotropic meshes, F. Alauzet, 2010 pdf) summarized as follows:

• The gradation on an edge ${\mathbf{e}}_{i,j}={\mathbf{x}}_j-{\mathbf{x}}_i$ is (with $a=l_{{\mathcal{M}}_i}(e_{i,j})/l_{{\mathcal{M}}_j}(e_{i,j})$ ) $$\max {\left(a,\frac{1}{a}\right)}^{\frac{1}{l_{\mathcal{M}}(e_{i,j})}}$$
• Given a metric $\mathcal{M}(\mathbf{x})=R^{T}diag(s_1^2,\cdots s_d^2)R$, is is possible to span a field at any location $\mathbf{y}$ as $${\mathcal{M}}_s(\mathbf{x},\mathbf{y})=R^{T}diag({\eta }_1^2{s}_1^2,\cdots {\eta }_d^2{s}_d^2)R$$
  with ${\eta }_i=1+s_i\Vert \mathbf{y}-\mathbf{x}{\Vert }_2\log (\beta )$ so that the gradation along $\mathbf{y}-\mathbf{x}$ is $\beta$
• In practice a maximum gradation is enforces along edge ${\mathbf{e}}_{i,j}={\mathbf{x}}_j-{\mathbf{x}}_i$, by modifying ${\mathcal{M}}_k$ to $$\widetilde{{\mathcal{M}}_j}={\mathcal{M}}_j\cap {\mathcal{M}}_s({\mathbf{x}}_i,{\mathbf{x}}_j)$$
  and $$\widetilde{{\mathcal{M}}_i}={\mathcal{M}}_i\cap {\mathcal{M}}_s({\mathbf{x}}_j,{\mathbf{x}}_i)$$
• Achieving a maximum gradation over all the edges in the mesh would require ${\#}_{edges}^2$ operations; in practices, the above operations is only applied on all the edges of the mesh a small number of times.

Thresholds on the local sizes

In order to ensure that the metric field will generate meshes with edge sizes between $h_{min}$ and $h_{max}$, the metric $\mathcal{M}=R^{T}diag(s_1,\cdots s_d)R$ with $R=({\mathbf{v}}_1|\cdots |{\mathbf{v}}_d)$ is modified as $$\mathcal{T}(\mathcal{M},h_{min},h_{max})=R^{T}diag(s_1,\cdots s_d)R$$ with ${\tilde{s}}_i=\min \left(\max \left(s_i,\frac{1}{h_{max}^2}\right),\frac{1}{h_{min}^2}\right)$.

Element-implied metric

Given an element $K$, defined by its vertices $({\mathbf{x}}_0,\cdots ,{\mathbf{x}}_d)$, the element-implied metric is the metric for which the element is equilateral

If $J_{K,\Delta }$ is the jacobian of the transformation from the reference equilateral element to the physical element, the element-implied metric is $${\mathcal{M}}_K=(J_{K,\Delta }{J}_{K,\Delta }^T{)}^{-1}$$

In practice $J_{K,\Delta }$ is computed as $J_{K,\Delta }=J_{K,\perp }{J}_{\perp ,\Delta }$ where - $J_{K,\perp }=({\mathbf{x}}_1-{\mathbf{x}}_0|\cdots |{\mathbf{x}}_d-{\mathbf{x}}_0)$ is the jacobian of transformation from the reference orthogonal element to the physical element - $J_{\perp ,\Delta }$ is the jacobian of transformation from the reference equilateral element to the reference orthogonal element (independent of $K$ )

Controling the step between two metrics

In the remeshing process, it may be interesting to limit the step between the element-implied and target metrics

Given two metric fields ${\mathcal{M}}_1$ and ${\mathcal{M}}_2$, the objective is to find $\mathcal{M}=\mathcal{L}({\mathcal{M}}_1,{\mathcal{M}}_2,f)$ as close as possible from ${\mathcal{M}}_2$ such that, for all edges $e$ $$1/f\le \sqrt{\frac{e^T\mathcal{M}e}{e^T{\mathcal{M}}_{1}e}}\le f$$ i.e. to have $$1/f\le {\lambda }_{min}({\mathcal{M}}_1^{-1/2}\mathcal{M}{\mathcal{M}}_1^{-1/2})\le {\lambda }_{max}({\mathcal{M}}_1^{-1/2}\mathcal{M}{\mathcal{M}}_1^{-1/2})\le f$$

• If as close as possible is defined as minimizing the Froebenius norm $\Vert {\mathcal{M}}_0^{-1/2}(\mathcal{M}-{\mathcal{M}}_2){\mathcal{M}}_0^{-1/2}{\Vert }_F$, the optimal $\mathcal{M}$ is computed as follows
  • compute ${\mathcal{N}}_1:={\mathcal{M}}_0^{-1/2}{\mathcal{M}}_1{\mathcal{M}}_0^{-1/2}$,
  • compute the eigenvalue decomposition $QD{Q}^T={\mathcal{N}}_1$,
  • compute ${\mathcal{N}}^{\ast }:=Q\mathrm{diag}({\hat{\lambda }}_i)Q^T$ where ${\hat{\lambda }}_i:=\min (\max ({\lambda }_i,1/f),f)$,
  • compute ${\mathcal{M}}^{\ast }:={\mathcal{M}}_0^{1/2}{\mathcal{N}}^{\ast }{\mathcal{M}}_0^{1/2}$.

(the thin blue lines represent $2{\mathcal{M}}_1$ and $1/2{\mathcal{M}}_1$)

See proof

Metric scaling

• The "ideal" metric $\mathcal{M}$ is modified using a series of constraints
  • Minimum and maximum edge sizes $h_{min}$ and $h_{max}$,
  • Maximum gradation $\beta$,
  • Maximum step $f$ with respect to the element implied metric ${\mathcal{M}}_i$,
  • A minimum curvature radius to edge lenth ratio $r$ (and optionally $y_{target}^{+}$), giving a fixed metric ${\mathcal{M}}_f$
• A scaling $\alpha$ is applied, such that the actual metric used for remeshing is $$\tilde{\mathcal{M}}=\mathcal{L}(\mathcal{T}(\alpha \mathcal{M},h_{min},h_{max})\cap {\mathcal{M}}_f,{\mathcal{M}}_i,f)$$
• $\alpha$ is chosen to have a given complexity $$\mathcal{C}(\tilde{\mathcal{M}})=\mathcal{N}$$
• A simple bisection is used to compute $\alpha$ if all the constraints can be met.

Maths

Notations

• One denotes the Frobenius norm of a matrix $A\in {\mathbb{R}}^{n\times n}$ by $\Vert A{\Vert }_F:=\mathrm{tr}(A^{T}A{)}^{1/2}=({\sum }_{i,j}{A}_{i,j}^2{)}^{1/2}$. This norm is associated with the scalar product $(A,B{)}_F:=\mathrm{tr}(A^{T}B)$.
• One denotes by $\le$ the partial order in the symmetric matrices space ${\mathcal{S}}_n(\mathbb{R})$: $S_1\le S_2$ if $S_2-S_1\in {\mathcal{S}}_n^{+}(\mathbb{R})$, i.e. for all $x\in {\mathbb{R}}^n$, $x^T{S}_{1}x\le x^T{S}_{2}x$.

Projection on a closed convex subset of the symmetric matrices

The following theorem has been proven in Computing a nearest symmetric positive semidefinite matrix, Higham Nicholas J, 1988:

Theorem 1 Let ${\mathcal{S}}_{\delta }:=\{S\in {\mathcal{S}}_n({\mathbb{R}}^n):\delta I_n\le S\}$. Then, ${\mathcal{S}}_{\delta }$ is a closed convex subset of the space ${\mathcal{S}}_n({\mathbb{R}}^n)$ and then there exists a projection application $P_{\delta }:{\mathcal{S}}_n({\mathbb{R}}^n)\to {\mathcal{S}}_{\delta }$ such that $$\forall A\in {\mathcal{S}}_n(\mathbb{R}),\Vert A-P_{\delta }(A){\Vert }_F=\inf \{\Vert A-S{\Vert }_F:S\in {\mathcal{S}}_{\delta }\}.>$$ Let $A\in {\mathcal{S}}_n(\mathbb{R})$ and $A=Q^{T}DQ$ its diagonalisation in an orthonormal basis: $Q\in {\mathcal{O}}_n(\mathbb{R})$, $D=\mathrm{diag}({\lambda }_i)$. Then, $$P_{\delta }(A)=Q\hat{D}{Q}^T,\text{where}\hat{D}:=\mathrm{diag}(\max ({\lambda }_i,\delta )).$$

Proof

Let $A\in {\mathcal{S}}_n(\mathbb{R})$ and its diagonalisation in a orthonomal basis $P^{T}AP=D=\mathrm{diag}({\lambda }_i)$. As $\Vert .{\Vert }_F$ is invariant by linear isometry ($\Vert PAQ{\Vert }_F^2=\mathrm{tr}((PAQ{)}^{T}PAQ)=\mathrm{tr}(Q^T{A}^T{P}^{T}PAQ)=\mathrm{tr}(Q{Q}^T{A}^{T}A)=\Vert A{\Vert }_F^2$, one gets $\Vert A-S{\Vert }_F^2=\Vert D-Q^{T}SQ{\Vert }_F^2$ and then the change of variable $S\in {\mathcal{S}}_n(\mathbb{R})\mapsto X=Q^{T}SQ\in {\mathcal{S}}_n(\mathbb{R})$ together with $\delta I\le S\iff \delta I\le Q^{T}SQ$, leads to the following problem $$\inf \{\Vert D-X{\Vert }_F^2:X\in {\mathcal{S}}_n(\mathbb{R}),\delta I\le X\}.$$ This problem can be computed explicitly. Since $X_{ii}=e_i^{T}X{e}_i\ge \delta e_i^T{e}_i=\delta$, ($e_i=({\delta }_{ij}{)}_j$ the canonical basis vectors) $$\Vert D-X{\Vert }_F^2=\sum _{i=1}^n({\lambda }_i-X_{ii}{)}^2+2\sum _{i<j}{X}_{ij}^2\ge \sum _{i=1}^n({\lambda }_i-X_{ii}{)}^2\ge \sum _{i:{\lambda }_i<\delta }({\lambda }_i-\delta {)}^2.$$ This lower bound is reached for the following diagonal matrix $X:=\hat{D}:=\mathrm{diag}({\hat{\lambda }}_i)$, where ${\hat{\lambda }}_i:=\max ({\lambda }_i,\delta )$, and moreover, $\hat{D}\ge \delta I$ and is in ${\mathcal{S}}_n(\mathbb{R})$, which ends the proof.

Generalization

One can generalize the Theorem above as follow:

Theorem 2 Given two symmetric real numbers $0<\alpha <\beta$, let ${\mathcal{S}}_{\alpha ,\beta }:=\{S\in {\mathcal{S}}_n(\mathbb{R}):\alpha I_n\le S\le \beta I_n\}$. ${\mathcal{S}}_{\alpha ,\beta }$ is a compact convex subset of ${\mathcal{S}}_n(\mathbb{R})$ and then let us denote $P_{\alpha ,\beta }:{\mathcal{S}}_n(\mathbb{R})\to {\mathcal{S}}_{\alpha ,\beta }$ the associated projection application: $$\forall A\in {\mathcal{S}}_n(\mathbb{R}),\Vert A-P_{\alpha ,\beta }(A){\Vert }_F=\inf \{\Vert A-S{\Vert }_F:S\in {\mathcal{S}}_{\alpha ,\beta }\}.>$$ Given $A\in {\mathcal{S}}_n(\mathbb{R})$ and $A=QD{Q}^T$ its diagonalisation in an orthonormal basis: $Q\in {\mathcal{O}}_n(\mathbb{R})$, $D=\mathrm{diag}({\lambda }_i)$. Then, $$P_{\alpha ,\beta }(A)=Q\hat{D}{Q}^T,\text{where}\hat{D}:=\mathrm{diag}({\hat{\lambda }}_i),{\hat{\lambda }}_i:=\min \left(\max \right({\lambda }_i,\alpha ),\beta ).$$

Proof

The proof follows the one above very closely. The lower bound is got using the inequality: $\alpha e_i^T{I}_n{e}_i\le e_i^{T}X{e}_i\le \beta e_i^T{I}_n{e}_i\Rightarrow \alpha \le X_{ii}\le \beta$: $$\Vert D-X{\Vert }_F^2\ge \sum _{i=1}^n({\lambda }_i-X_{ii}{)}^2+2\sum _{i<j}{X}_{ij}^2$$ $$\ge \sum _{i=1}^n({\lambda }_i-X_{ii}{)}^2$$ $$\ge \sum _{i:{\lambda }_i<\alpha }({\lambda }_i-\alpha {)}^2+\sum _{i:{\lambda }_i>\beta }({\lambda }_i-\beta {)}^2,$$ which is reached for $\hat{D}:=\mathrm{diag}({\hat{\lambda }}_i)$ where ${\hat{\lambda }}_i=\alpha$ if ${\lambda }_i<\alpha$, ${\hat{\lambda }}_i=\beta$ if ${\lambda }_i>\beta$ and otherwise ${\hat{\lambda }}_i={\lambda }_i$. Again, the matrix $\hat{D}$ is symmetric and indeed in ${\mathcal{S}}_{\alpha ,\beta }$ (because since $\hat{D}$ is diagonal, $\alpha \le {\hat{\lambda }}_i\le \beta \iff \alpha I_n\le \hat{D}\le \beta I_n$) and thus is the unique solution of the problem (existence and uniqueness beeing given by the projection on closed convex sets in a Hilbert space).

Metric optimization problem

Given two metrics ${\mathcal{M}}_1,{\mathcal{M}}_2\in {\mathcal{S}}_n^{++}(\mathbb{R})$. The deformation ratio between a metric $\mathcal{M}\in S_n^{++}(\mathbb{R})$ and ${\mathcal{M}}_0$ is defined as a function of x∈Rn</span>{0} r(M,M0​;x):=(xTM0​xxTMx​)1/2.

The problem.

Given two metrics ${\mathcal{M}}_1,{\mathcal{M}}_2\in {\mathcal{S}}_n^{++}(\mathbb{R})$, and two real numbers $0<\alpha <\beta$, let ${\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)$ be the set of metrics which are not to stretched with regard to ${\mathcal{M}}_0$: Rα,β​(M0​):={M∈Sn​(R):∀x∈Rn</span>{0},α≤r(M,M0​;x)2≤β}. Again, this set is compact and convex. Given a matrix norm $$|||.|||$$, we want to find $${\mathcal{M}}^{\ast }\in {\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)$$ which minimize the distance to $${\mathcal{M}}_1$$, i.e. find $${\mathcal{M}}^{\ast }\in {\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)$$ such that: $$|||{\mathcal{M}}_1-{\mathcal{M}}^{\ast }|||=\inf \{|||{\mathcal{M}}_1-\mathcal{M}|||:\mathcal{M}\in {\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)\}$$

lemma : Let $\mathcal{M},{\mathcal{M}}_0\in {\mathcal{S}}_n^{++}(\mathbb{R})$ two metrics > and $\alpha ,\beta \in \mathbb{R}$. The function x∈Rn</span>{0}↦r(M,M0​;x)2 is bounded and reaches its bounds which are the minimum and maximum of the eigenvalues of M0>​−1/2MM0−1/2​. Thus, >∀x∈Rn</span>{0},>α≤r(M,M0​;x)≤β> ⇔>αIn​≤M0−1/2​MM0−1/2​≤β>In​> ⇔>αM0​≤M≤βM0​>

Thanks to this lemma, the admissible set ${\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)$ can be re-writen as $${\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)=\{\mathcal{M}\in {\mathcal{S}}_n(\mathbb{R}):\alpha {\mathcal{M}}_0\le \mathcal{M}\le \beta {\mathcal{M}}_0\}$$ $$=\{\mathcal{M}\in {\mathcal{S}}_n(\mathbb{R}):\alpha I_n\le {\mathcal{M}}_0^{-1/2}\mathcal{M}{\mathcal{M}}_0^{-1/2}\le \beta I_n\}.$$

In the ${\mathcal{M}}_0$-twisted Frobenius norm

First, one studies the case of choosing $$|||\mathcal{M}|||:=\Vert {\mathcal{M}}_1-\mathcal{M}{\Vert }_{{\mathcal{M}}_0,F}:=\Vert {\mathcal{M}}_0^{-1/2}\mathcal{M}{\mathcal{M}}_0^{-1/2}{\Vert }_F.$$ The above problem reads as follows. Find ${\mathcal{M}}^{\ast }\in {\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)$ such that: $$|||{\mathcal{M}}_1-{\mathcal{M}}^{\ast }|||=\inf \{||{\mathcal{M}}_0^{-1/2}({\mathcal{M}}_1-\mathcal{M}){\mathcal{M}}_0^{-1/2}|{|}_F:\mathcal{M}\in {\mathcal{S}}_n(\mathbb{R}),{\mathcal{M}}_0^{-1/2}\mathcal{M}{\mathcal{M}}_0^{-1/2}\in {\mathcal{S}}_{\alpha ,\beta }\}.$$ It is then clear that the change of variable $\mathcal{M}\in {\mathcal{S}}_n(\mathbb{R})\to \mathcal{N}:={\mathcal{M}}_0^{-1/2}\mathcal{M}{\mathcal{M}}_0^{-1/2}\in {\mathcal{S}}_n(\mathbb{R})$ leads to a problem of the form of Theorem 2. Denoting ${\mathcal{N}}_1:={\mathcal{M}}_0^{-1/2}{\mathcal{M}}_1{\mathcal{M}}_0^{-1/2}$ the problem reads find ${\mathcal{M}}^{\ast }={\mathcal{M}}_0^{1/2}{\mathcal{N}}^{\ast }{\mathcal{M}}_0^{1/2}\in {\mathcal{R}}_{\alpha ,\beta }({\mathcal{M}}_0)$ such that $$\Vert {\mathcal{N}}_1-{\mathcal{N}}^{\ast }{\Vert }_F=\inf \{\Vert {\mathcal{N}}_1-\mathcal{N}{\Vert }_F:\mathcal{N}\in {\mathcal{S}}_{\alpha ,\beta }\}.$$ Then, from Theorem 2, the solution is $${\mathcal{M}}^{\ast }={\mathcal{M}}_0^{1/2}{P}_{\alpha ,\beta }({\mathcal{N}}_1){\mathcal{M}}_0^{1/2}$$ and can be computed as follows:

• Compute ${\mathcal{N}}_1:={\mathcal{M}}_0^{-1/2}{\mathcal{M}}_1{\mathcal{M}}_0^{-1/2}$,
• Find $Q\in {\mathcal{O}}_n(\mathbb{R})\text{ and }D=\mathrm{diag}({\lambda }_i)$ such that $QD{Q}^T={\mathcal{N}}_1$.
• Compute ${\mathcal{N}}^{\ast }:=P_{\alpha ,\beta }({\mathcal{N}}_1)=Q\mathrm{diag}({\hat{\lambda }}_i)Q^T,\text{where }{\hat{\lambda }}_i:=\min \left(\max \right({\lambda }_i,\alpha ),\beta )$.
• Compute ${\mathcal{M}}^{\ast }:={\mathcal{M}}_0^{1/2}{\mathcal{N}}^{\ast }{\mathcal{M}}_0^{1/2}$

Remark Maybe a more suitable formulation would minimizing the Frobenius norm $|||{\mathcal{M}}_1-\mathcal{M}|||:=\Vert {\mathcal{M}}_1-\mathcal{M}{\Vert }_F$. This leads to a different problem. For instance, if $${\mathcal{M}}_0=\left(\begin{array}{cc}1& 0\\ 0& 1/\epsilon \end{array}\right)$$ where $0<\epsilon \ll 1$. One gets, denoting the error $\mathcal{E}:=(e_{ij}):={\mathcal{M}}_1-{\mathcal{M}}^{\ast }$ $$\Vert \mathcal{E}{\Vert }_{{\mathcal{M}}_0,F}^2=e_{11}^2+2\epsilon e_{12}^2+{\epsilon }^2{e}_{22}^2,$$ instead of $$\Vert \mathcal{E}{\Vert }_F^2=e_{11}^2+2e_{12}^2+e_{22}^2.$$ The two distance measures give very different results... does it matter in practice?

Intersection of metrics

Problem.

Given two metrics ${\mathcal{M}}_1,{\mathcal{M}}_2\in {\mathcal{S}}_n^{++}(\mathbb{R})$, we want to find a metric ${\mathcal{M}}_{1\cap 2}\in {\mathcal{S}}_n^{++}(\mathbb{R})$ which respects the most restrictive length of ${\mathcal{M}}_1$ and ${\mathcal{M}}_1$. Moreover, we want that this metric beeing maximal. More formally, the lenght restriction can be expressed by ∀x∈Rn</span>{0},xTM1∩2​x≤xTM1​xandxTM1∩2​x≤xTM2​x, indeed, $$\sqrt{x^T\mathcal{M}x}$$ is the length of $$x$$ associated with a metric $$\mathcal{M}$$. The local volume associated with a metric $$\mathcal{M}$$ is its determinant: $$\det \mathcal{M}$$, so that the optimization problem reads $$\max \det {\mathcal{M}}_{1\cap 2}$$ such that $${\mathcal{M}}_{1\cap 2}\in {\mathcal{S}}_n(\mathbb{R}),$$ $$0\le {\mathcal{M}}_{1\cap 2}\le {\mathcal{M}}_1,$$ $$0\le {\mathcal{M}}_{1\cap 2}\le {\mathcal{M}}_2.$$

Solution of the problem.

First, one multiplies by ${\mathcal{M}}_1^{-1/2}\in {\mathcal{S}}_n^{++}(\mathbb{R})$ on each side of the constraints, which leads to $$0\le {\mathcal{M}}_1^{-1/2}{\mathcal{M}}_{1\cap 2}{\mathcal{M}}_1^{-1/2}\le I_n$$ $$0\le {\mathcal{M}}_1^{-1/2}{\mathcal{M}}_{1\cap 2}{\mathcal{M}}_1^{-1/2}\le {\mathcal{M}}_1^{-1/2}{\mathcal{M}}_2{\mathcal{M}}_1^{-1/2}.$$ Then, denoting ${\overline{\mathcal{M}}}_2:={\mathcal{M}}_1^{-1/2}{\mathcal{M}}_2{\mathcal{M}}_1^{-1/2}$ and applying the change of variable $\mathcal{M}\mapsto \mathcal{N}:={\mathcal{M}}_1^{-1/2}\mathcal{M}{\mathcal{M}}_1^{-1/2}$, one gets the simplified problem: $$\max \det \left({\mathcal{M}}_1^{1/2}\mathcal{N}{\mathcal{M}}_1^{1/2}\right)$$ such that $$\mathcal{N}\in {\mathcal{S}}_n(\mathbb{R}),$$ $$0\le \mathcal{N}\le I_n,$$ $$0\le \mathcal{N}\le {\overline{\mathcal{M}}}_2.$$

This problem can be even more simplified using $\det \left({\mathcal{M}}_1^{1/2}\mathcal{N}{\mathcal{M}}_1^{1/2}\right)=\det {\mathcal{M}}_1\det \mathcal{N}$ (does not change the $\max$ objective as $\det {\mathcal{M}}_1>0$) and by diagonalization of ${\overline{\mathcal{M}}}_2$ in an orthonomal basis: $${\overline{\mathcal{M}}}_2=P^{T}DP,\text{with }{P}^{T}P=I_n.$$ Indeed, $0\le \mathcal{N}\le P^{T}DP\Rightarrow 0\le P\mathcal{N}{P}^T\le D$ and $0\le \mathcal{N}\le I_n\Rightarrow 0\le P\mathcal{N}{P}^T\le I_n$, and let us change again the variable $\mathcal{X}:=P\mathcal{N}{P}^T$, with $\det \mathcal{N}=\det \mathcal{X}$, the problem is finally: $$\max \det \mathcal{X}$$ such that $$\mathcal{X}\in {\mathcal{S}}_n(\mathbb{R}),$$ $$0\le \mathcal{X}\le I_n,$$ $$0\le \mathcal{X}\le D.$$

This problem can now to be solved explicitly by finding an upper bound. Let $\mathcal{Y}\in {\mathcal{S}}_n(\mathbb{R})$ such that $\mathcal{X}={\mathcal{Y}}^2$, then Hadamard inequality provides $\det \mathcal{X}=(\det \mathcal{Y}{)}^2\le {\prod }_j|{\mathcal{Y}}_{\bullet i}{|}_2^2$. Moreover, ${\mathcal{X}}_{jj}={\sum }_k{\mathcal{Y}}_{jk}{\mathcal{Y}}_{kj}=|{\mathcal{Y}}_{\bullet j}{|}_2^2$, from which $\det \mathcal{X}\le {\prod }_i{\mathcal{X}}_{ii}$ and then, using the two constraints: $0\le \mathcal{X}\le I_n\text{ and }0\le \mathcal{X}\le D\Rightarrow 0\le {\mathcal{X}}_{ii}\le \min (D_{ii},1)$, one gets the upper bound $$\det \mathcal{X}\le \prod _{i=1}^n{\mathcal{X}}_{ii}\le \prod _{i=1}^n\min (D_{ii},1),$$ reached for the diagonal matrix $D^{\ast }:=\mathrm{diag}(\min (D_{ii},1))$. Moreover, this matrix is symmetric and verifies $0\le D^{\ast }\le I_n$ and $0\le D^{\ast }\le D$.

Finally, applying the two changes of variables, the unique solution of the problem is $${\mathcal{M}}_{1\cap 2}={\mathcal{M}}_1^{1/2}{P}^T{D}^{\ast }P{\mathcal{M}}_1^{1/2}$$

Hadamard inequality

Given a matrix $M=[v_1|\dots |v_n]\in {\mathbb{C}}^{n\times n}$, one has $$\left|\det M\right|\le \prod _{i=1}^n|v_i{|}_2,$$ with equality iff the $(v_1,\dots ,v_n)$ are orthogonal.

The result if clear if $M$ is singular, if not let $\hat{M}:=[{\hat{v}}_1|\dots |{\hat{v}}_n]$, ${\hat{v}}_i=v_i/|v_i{|}_2$ so that one has to show that $|\det \hat{M}|\le 1$. ${\hat{M}}^{\ast }\hat{M}\in {\mathcal{H}}_n(\mathbb{C})$ is hermitian hence diagonalizable in an orthonomal basis, denoting $({\lambda }_1,\dots ,{\lambda }_n)$ its eigenvalues, one has $$|\det \hat{M}{|}^2=\det {\hat{M}}^{\ast }\hat{M}=\prod _{i=1}^n{\lambda }_i$$ $$\le {\left(\frac{1}{n}\sum _{i=1}^n{\lambda }_i\right)}^n={\left(\frac{1}{n}\mathrm{tr}\left({\hat{M}}^{\ast }M\right)\right)}^n={\left(\frac{1}{n}\sum _{i=1}^n\sum _{k=1}^n\overline{{\hat{M}}_{ki}}{\hat{M}}_{ki}\right)}^n$$ $$={\left(\frac{1}{n}\sum _{i=1}^n\sum _{k=1}^n|{\hat{M}}_{ki}{|}^2\right)}^n={\left(\frac{1}{n}\sum _{i=1}^n|{\hat{v}}_i{|}_2^2\right)}^n$$