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You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $B^i = \mu^i_{\ \ j} H^j$, then the tensor components are related via

$$\tilde B_{ij} = \epsilon_{ijk} B^k = \epsilon_{ijk} \mu^k_{\ \ \ell} H^\ell = \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n} \tilde H_{mn}$$ $$\implies B_{ij} = \tilde \mu_{ij}^{\ \ mn} H_{mn}$$ where $$\tilde \mu_{ij}^{\ \ mn}\equiv \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n}$$ Note that $\tilde \mu$ is antisymmetric in each pair of indices, and therefore has nine independent components - just as one would expect, since $\mu$ is a generic $3\times 3$ matrix. Generalizing to $n$-dimensions, $\tilde \mu$ would have $d^2(d-1)^2/4$ independent components, and would express a general proportional relationship between two $(0,2)$-tensor fields.

To further clarify the summation convention, consider the following example. The component$\tilde \mu_{12}^{\ \ \ 12}$ is

$$\tilde \mu_{12}^{\ \ \ 12} = \sum_{\ell=1}^3 \sum_{k=1}^3\frac{1}{2}\epsilon_{12k} \mu^k_{\ \ \ell} \epsilon^{\ell 12}$$ Since $\epsilon$ is completely antisymmetric, there is only one nonzero term, namely $k=3$ (otherwise $\epsilon_{12k}=0$) and $\ell=3$ (otherwise $\epsilon^{\ell 12}=0$). Therefore,

$$\tilde \mu_{12}^{\ \ \ 12} = \frac{1}{2}\underbrace{\epsilon_{123}}_{=1} \ \mu^3_{\ \ 3}\ \underbrace{\epsilon^{312}}_{=1} = \frac{1}{2} \mu^3_{\ \ 3}$$

You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $B^i = \mu^i_{\ \ j} H^j$, then the tensor components are related via

$$\tilde B_{ij} = \epsilon_{ijk} B^k = \epsilon_{ijk} \mu^k_{\ \ \ell} H^\ell = \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n} \tilde H_{mn}$$ $$\implies B_{ij} = \tilde \mu_{ij}^{\ \ mn} H_{mn}$$ where $$\tilde \mu_{ij}^{\ \ mn}\equiv \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n}$$ Note that $\tilde \mu$ is antisymmetric in each pair of indices, and therefore has nine independent components - just as one would expect, since $\mu$ is a generic $3\times 3$ matrix. Generalizing to $n$-dimensions, $\tilde \mu$ would have $d^2(d-1)^2/4$ independent components, and would express a general proportional relationship between two $(0,2)$-tensor fields.

It's also important to note that the $B\leftrightarrow \tilde B$ correspondence only holds in $3$ dimensions; in higher (resp. lower) dimensions, the magnetic field $\tilde B$ has too many (resp. too few) components to be represented as a (pseudo)vector. In the same way, the correspondence between the $(2,2)$-tensor $\tilde \mu$ and a $(1,1)$-tensor $\mu$ is also a feature of $3$ dimensions only. The magnetic field is really a $(0,2)$-tensor $\tilde B$ which can be associated to a pseudovector $B$ in the special case that $d=3$; similarly the magnetic susceptibility is really a $(2,2)$-tensor $\tilde \mu$ which can be associated to a $(1,1)$-tensor $\mu$ only when $d=3$.

To further clarify the summation convention, consider the following example. The component  $\tilde \mu_{12}^{\ \ \ 12}$ is

$$\tilde \mu_{12}^{\ \ \ 12} = \sum_{\ell=1}^3 \sum_{k=1}^3\frac{1}{2}\epsilon_{12k} \mu^k_{\ \ \ell} \epsilon^{\ell 12}$$ Since $\epsilon$ is completely antisymmetric, there is only one nonzero term, namely $k=3$ (otherwise $\epsilon_{12k}=0$) and $\ell=3$ (otherwise $\epsilon^{\ell 12}=0$). Therefore,

$$\tilde \mu_{12}^{\ \ \ 12} = \frac{1}{2}\underbrace{\epsilon_{123}}_{=1} \ \mu^3_{\ \ 3}\ \underbrace{\epsilon^{312}}_{=1} = \frac{1}{2} \mu^3_{\ \ 3}$$

You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $B^i = \mu^i_{\ \ j} H^j$, then the tensor components are related via

$$\tilde B_{ij} = \epsilon_{ijk} B^k = \epsilon_{ijk} \mu^k_{\ \ \ell} H^\ell = \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n} \tilde H_{mn}$$ $$\implies B_{ij} = \tilde \mu_{ij}^{\ \ mn} H_{mn}$$ where $$\tilde \mu_{ij}^{\ \ mn}\equiv \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n}$$ Note that $\tilde \mu$ is antisymmetric in each pair of indices, and therefore has nine independent components - just as one would expect, since $\mu$ is a generic $3\times 3$ matrix. Generalizing to $n$-dimensions, $\tilde \mu$ would have $d^2(d-1)^2/4$ independent components, and would express a general proportional relationship between two $(0,2)$-tensor fields.

You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $B^i = \mu^i_{\ \ j} H^j$, then the tensor components are related via

$$\tilde B_{ij} = \epsilon_{ijk} B^k = \epsilon_{ijk} \mu^k_{\ \ \ell} H^\ell = \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n} \tilde H_{mn}$$ $$\implies B_{ij} = \tilde \mu_{ij}^{\ \ mn} H_{mn}$$ where $$\tilde \mu_{ij}^{\ \ mn}\equiv \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n}$$ Note that $\tilde \mu$ is antisymmetric in each pair of indices, and therefore has nine independent components - just as one would expect, since $\mu$ is a generic $3\times 3$ matrix. Generalizing to $n$-dimensions, $\tilde \mu$ would have $d^2(d-1)^2/4$ independent components, and would express a general proportional relationship between two $(0,2)$-tensor fields.

To further clarify the summation convention, consider the following example. The component$\tilde \mu_{12}^{\ \ \ 12}$ is

$$\tilde \mu_{12}^{\ \ \ 12} = \sum_{\ell=1}^3 \sum_{k=1}^3\frac{1}{2}\epsilon_{12k} \mu^k_{\ \ \ell} \epsilon^{\ell 12}$$ Since $\epsilon$ is completely antisymmetric, there is only one nonzero term, namely $k=3$ (otherwise $\epsilon_{12k}=0$) and $\ell=3$ (otherwise $\epsilon^{\ell 12}=0$). Therefore,

$$\tilde \mu_{12}^{\ \ \ 12} = \frac{1}{2}\underbrace{\epsilon_{123}}_{=1} \ \mu^3_{\ \ 3}\ \underbrace{\epsilon^{312}}_{=1} = \frac{1}{2} \mu^3_{\ \ 3}$$

You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $H^i = \mu^i_{\ \ j} B^j$, then the tensor components are related via

$$\tilde H_{ij} = \epsilon_{ijk} H^k = \epsilon_{ijk} \mu^k_{\ \ \ell} B^\ell = \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n} \tilde B_{mn}$$ $$\implies H_{ij} = \tilde \mu_{ij}^{\ \ mn} B_{mn}$$ where $$\tilde \mu_{ij}^{\ \ mn}\equiv \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n}$$ Note that $\tilde \mu$ is antisymmetric in each pair of indices, and therefore has nine independent components - just as one would expect, since $\mu$ is a generic $3\times 3$ matrix. Generalizing to $n$-dimensions, $\tilde \mu$ would have $d^2(d-1)^2/4$ independent components, and would express a general proportional relationship between two $(0,2)$-tensor fields.

You already have $\tilde B_{ij} = \epsilon_{ijk}B^k \iff B^k = \frac{1}{2}\epsilon^{ijk}\tilde B_{ij}$ and similarly for $H$ and $\tilde H$. If the pseudovector components are related via $B^i = \mu^i_{\ \ j} H^j$, then the tensor components are related via

$$\tilde B_{ij} = \epsilon_{ijk} B^k = \epsilon_{ijk} \mu^k_{\ \ \ell} H^\ell = \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n} \tilde H_{mn}$$ $$\implies B_{ij} = \tilde \mu_{ij}^{\ \ mn} H_{mn}$$ where $$\tilde \mu_{ij}^{\ \ mn}\equiv \frac{1}{2}\epsilon_{ijk} \mu^k_{\ \ \ell} \epsilon^{\ell m n}$$ Note that $\tilde \mu$ is antisymmetric in each pair of indices, and therefore has nine independent components - just as one would expect, since $\mu$ is a generic $3\times 3$ matrix. Generalizing to $n$-dimensions, $\tilde \mu$ would have $d^2(d-1)^2/4$ independent components, and would express a general proportional relationship between two $(0,2)$-tensor fields.