$\nu^{\omega}:=f^{\omega}d\mu^{\omega}$, and let $\nu_{L}$ denote the Loeb
measure associated to $\nu^{\omega}$.

Then, $\nu_{L}$ also pushes forward to a Radon measure $\nu$ on $\hat{X}$, and
$\nu\ll\mu$.

###### ?proofname?.

Let $f^{\omega}$ be an $S$-integrable function with respect to $\mu^{\omega}$,
with $\nu^{\omega}(A)=\int_{A}f^{\omega}d\mu^{\omega}$ for every internal
Borel set $A\in\mathcal{B}^{\omega}$. Then,

$\nu_{L}(A)=st\left(\int_{A}f^{\omega}d\mu^{\omega}\right)=\int_{A}st\circ
f^{\omega}d\mu_{L}$

and also, for all $B\in\sigma(\mathcal{B}^{\omega})$,

$\nu_{L}(B)=\int_{B}(st\circ f^{\omega})d\mu_{L}.$

Let $\mu:=\mu_{L}\circ st_{\hat{d}}^{-1}$. By assumption, $\mu$ has separable
support. Let $\mathbb{E}_{st_{\hat{d}}^{-1}}$ denote the conditional
expectation of $st\circ f^{\omega}$ with respect to the sub-$\sigma$-algebra
of $\sigma(\mathcal{B}^{\omega})$ formed by taking the $st_{\hat{d}}$-preimage
of the Borel $\sigma$-algebra on $\hat{X}$ restricted to $supp(\mu)$. (That
this is indeed a sub-$\sigma$-algebra follows from Lemma 15.) Note that for
all Borel $\tilde{B}\subseteq supp(\mu)$,

$\nu_{L}(st_{\hat{d}}^{-1}(\tilde{B}))=\int_{st_{\hat{d}}^{-1}(\tilde{B})}(st\circ
f^{\omega})d\mu_{L}=\int_{st_{\hat{d}}^{-1}(\tilde{B})}\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ
f^{\omega})d\mu_{L}.$

Since $\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ f^{\omega})$ is constant on
fibers $st_{\hat{d}}^{-1}(x)$ for $\mu_{L}$-almost all $x$, we see that
$\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ f^{\omega})\circ
st_{\hat{d}}^{-1}:supp(\mu)\rightarrow\mathbb{R}$ is a well-defined Borel
measurable function, and change-of-variable for measures shows that

$\displaystyle\int_{st_{\hat{d}}^{-1}(\tilde{B})}\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ
f^{\omega})d\mu_{L}$
$\displaystyle=\int_{\tilde{B}}\left(\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ
f^{\omega})\circ st_{\hat{d}}^{-1}\right)d(\mu_{L}\circ st_{\hat{d}}^{-1})$
$\displaystyle=\int_{\tilde{B}}\left(\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ
f^{\omega})\circ st_{\hat{d}}^{-1}\right)d\mu.$

Hence, for all Borel $\tilde{B}\subseteq supp(\mu)$,

$(\nu_{L}\circ
st_{\hat{d}}^{-1})(\tilde{B})=\int_{\tilde{B}}\left(\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ
f^{\omega})\circ st_{\hat{d}}^{-1}\right)d\mu.$

Hence, if $\nu^{\omega}\ll^{\omega}\mu^{\omega}$ with “internal Radon-Nikodym
derivative” $f^{\omega}$, and $f^{\omega}$ is $S$-integrable with respect to
$\mu^{\omega}$, and $\mu_{L}$ pushes forward to a Radon measure $\mu$ on
$\hat{X}$, it follows that $\nu_{L}$ also pushes forward to a Radon measure
$\nu$ on $\hat{X}$, and $\nu\ll\mu$ with
$\frac{d\nu}{d\mu}=\mathbb{E}_{st_{\hat{d}}^{-1}}(st\circ f^{\omega})\circ
st_{\hat{d}}^{-1}$. ∎

The next two lemmas are not stated using sharp assumptions, but rather using
assumptions which are simply strong enough to run the argument for Theorem 28.

######  Lemma 32.

(Stability of integral functionals under lifting) Let
$\mathcal{F}(\nu):\mathcal{P}(\hat{X})\rightarrow\mathbb{R}$ be an integral
functional of the form

$\mathcal{F}(\nu)=\begin{cases}\int_{\hat{X}}\varphi\left(\frac{d\nu}{d\mu}\right)d\mu&\nu\ll\mu\\\
\infty&\nu\not\ll\mu\end{cases}$

where $\varphi:\mathbb{R}_{+}\rightarrow\mathbb{R}$ is some continuous
function such that $\varphi(x)\leq cx^{\alpha}+d$ for constants
$c,d,\alpha>0$. Let $\varphi^{\omega}:=[(\varphi)]$ be the hyperreal extension
of $\varphi$, let $\mu^{\omega}$ be an internal probability measure whose
associated Loeb measure $\mu_{L}$ pushes forward to a Radon probability
measure $\mu$ on $\hat{X}$, and let $\nu\ll\mu$. Suppose that
$\frac{d\nu}{d\mu}\in L^{q}(\mu)$ where $q>\alpha$. There exists an
$S$-integrable lifting of $\frac{d\nu}{d\mu}\circ st_{\hat{d}}$, denoted
$H^{\omega}$, and for any such lifting,

$\mathcal{F}(\nu)\approx\int_{X^{\omega}}\varphi^{\omega}\left(H^{\omega}\right)d\mu^{\omega}.$

###### ?proofname?.

Let $h(x)=\frac{d\nu}{d\mu}(x)$. By change of variables, it follows from the
fact that $\mu_{L}\circ st_{\hat{d}}^{-1}=\mu$ that

$\int_{\hat{X}}\varphi\left(h(x)\right)d\mu(x)=\int_{X^{\omega}}\varphi\left(h(st_{\hat{d}}(y))\right)d\mu_{L}(y).$

In particular the function $h\circ st_{\hat{d}}$ is $\mu_{L}$-measurable, so
there exists an internal function
$H^{\omega}:X^{\omega}\rightarrow\mathbb{R}^{\omega}$ which is an
$S$-integrable lifting of $h\circ st_{\hat{d}}$. In particular
$H^{\omega}\approx h\circ st_{\hat{d}}$ $\mu_{L}$-almost surely. Moreover,
since $h\in L^{q}(\mu)$, it follows by change of variables that $h\circ
st_{\hat{d}}\in L^{q}(\mu_{L})$, and therefore
$st\left(\int|H^{\omega}|^{p}d\mu^{\omega}\right)<\infty$ by Fact 7. At the
same time, $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ is continuous, so
$\varphi^{\omega}$ has the property that $\varphi^{\omega}(x)=\varphi(x)$ for
all (standard) $x\in\mathbb{R}$, and additionally
$\varphi^{\omega}(x)\approx\varphi^{\omega}(y)$ whenever $x\approx y$ [24,
Theorem 7.1.1]. Therefore, for $\mu_{L}$-almost all $y$,

$\varphi\left(h(st(y))\right)=st(\varphi^{\omega}(H^{\omega}(y))).$

It follows that

$\int_{X^{\omega}}\varphi\left(h(st_{\hat{d}}(y))\right)d\mu_{L}(y)=\int_{X^{\omega}}st\left(\varphi^{\omega}(H^{\omega}(y))\right)d\mu_{L}(y).$

Lastly, let us check that $\varphi^{\omega}\circ H^{\omega}$ is
$S$-integrable. Using our polynomial bound on $\varphi$, we deduce that

$\varphi^{\omega}\circ H^{\omega}\leq c(H^{\omega})^{\alpha}+d.$

It follows that $st\left(\int|\varphi^{\omega}\circ
H^{\omega}|^{q/\alpha}d\mu^{\omega}\right)<\infty$, which implies (thanks to
Fact 8) that since $q/\alpha>1$, that $\varphi\circ H^{\omega}$ is
$S$-integrable.

Finally, it follows from $S$-integrability of
$\varphi^{\omega}(H^{\omega}(y))$, and the fact that $\mu_{L}$ is a finite
measure, that

$\int_{X^{\omega}}st\left(\varphi^{\omega}(H^{\omega}(y))\right)d\mu_{L}(y)\approx\int_{X^{\omega}}\varphi^{\omega}(H^{\omega}(y))d\mu^{\omega}(y).$

Hence
$\int_{\hat{X}}\varphi\left(h(x)\right)d\mu(x)\approx\int_{X^{\omega}}\varphi^{\omega}(H^{\omega}(y))d\mu^{\omega}(y)$
as desired. ∎

We also have a sort of converse:

######  Lemma 33.

(Stability of internal integral functionals under pushdown) Let
$\varphi:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ be a continuous, _convex_
function, such that $\varphi(x)\leq cx^{\alpha}+d$ for constants
$c,d,\alpha>0$. Let $\varphi^{\omega}:=[(\varphi)]$ denote the hyperreal
extension of $\varphi$. Let $\mu^{\omega}\in\mathcal{P}(X)^{\omega}$ be an
internal probability measure whose associated Loeb measure $\mu_{L}$ pushes
forward to a Radon probability measure $\mu$ on $\hat{X}$. Define

$\mathcal{F}^{\omega}(\nu^{\omega}):=\begin{cases}\int\varphi^{\omega}\circ
f^{\omega}d\mu^{\omega}&\nu^{\omega}\ll^{\omega}\mu^{\omega};\nu^{\omega}=f^{\omega}d\mu^{\omega}\\\
\infty^{\omega}&\text{otherwise}.\end{cases}$

(Here $\infty^{\omega}$ denotes a point formally added to
$\mathbb{R}^{\omega}$ such that $x^{\omega}<\infty^{\omega}$ for all
$x^{\omega}\in\mathbb{R}^{\omega}$.)

Then, if $st\left(\int|f^{\omega}|^{q}d\mu^{\omega}\right)<\infty$ for some
$q>\max\\{1,\alpha\\}$, it follows that $\nu_{L}:=\text{Loeb}(\nu^{\omega})$
pushes forward to some Radon probability measure $\nu$ on $\hat{X}$, and
$\nu\ll\mu$; and

$\mathcal{F}(\nu):=\int_{\hat{X}}\varphi\left(\frac{d\nu}{d\mu}\right)d\mu\lessapprox\mathcal{F}^{\omega}(\nu^{\omega}).$

###### ?proofname?.

Suppose that $\nu^{\omega}=f^{\omega}d\mu^{\omega}$, and
$st\left(\int|f^{\omega}|^{q}d\mu^{\omega}\right)<\infty$ for some $q>1$ (and
therefore $f^{\omega}$ is $S$-integrable w.r.t. $\mu^{\omega}$, by Fact 8),
and that $\mu_{L}\circ st_{\hat{d}}^{-1}$ is well-defined and a Radon
probability measure on $\hat{X}$. By Lemma 31, it follows that
$\nu:=\nu_{L}\circ st_{\hat{d}}^{-1}$ is also well-defined and a Radon
probability measure on $\hat{X}$, and $\nu\ll\mu$.

Since $\varphi^{\omega}\circ f^{\omega}\leq c(f^{\omega})^{\alpha}+d$, it
follows, as in the proof of Lemma 32 that $\varphi^{\omega}\circ f^{\omega}\in
SL^{q/\alpha}(\mu^{\omega})$ with $q/\alpha>1$, and in particular
$\varphi^{\omega}\circ f^{\omega}$ is $S$-integrable.

Therefore,

$\displaystyle st\left(\int\varphi^{\omega}\circ
f^{\omega}d\mu^{\omega}\right)$ $\displaystyle=\int st(\varphi^{\omega}\circ
f^{\omega})d\mu_{L}$ $\displaystyle=\int\varphi\circ st(f^{\omega})d\mu_{L}.$

Now, $\varphi\circ st(f^{\omega})$ is
$\sigma(\mathcal{B}^{\omega})$-measurable. As before, in Lemma 31, we take the
conditional expectation $\mathbb{E}_{st_{\hat{d}}^{-1}}\varphi\circ
st(f^{\omega})$; since $\varphi$ is convex, Jensen’s inequality for
conditional expectations [11, Proposition 10.1.9] shows that
$\varphi\circ\mathbb{E}_{st_{\hat{d}}^{-1}}st(f^{\omega})\leq\mathbb{E}_{st_{\hat{d}}^{-1}}\varphi\circ
st(f^{\omega})$ except on a set of $\mu_{L}$-measure zero. Thus,

$\displaystyle\int\varphi\circ\mathbb{E}_{st_{\hat{d}}^{-1}}st(f^{\omega})d\mu_{L}$
$\displaystyle\leq\int\mathbb{E}_{st_{\hat{d}}^{-1}}\varphi\circ
st(f^{\omega})d\mu_{L}$ $\displaystyle=\int\varphi\circ
st(f^{\omega})d\mu_{L}$ $\displaystyle=\int st(\varphi^{\omega}\circ
f^{\omega})d\mu_{L}$ $\displaystyle\approx\mathcal{F}^{\omega}(\nu^{\omega}).$

Performing a change of variables, as in Lemma 31, we see

$\int_{X^{\omega}}\varphi\circ\mathbb{E}_{st_{\hat{d}}^{-1}}st(f^{\omega})d\mu_{L}=\int_{\hat{X}}\varphi\circ\left(\mathbb{E}_{st_{\hat{d}}^{-1}}st(f^{\omega})\circ
st_{\hat{d}}^{-1}\right)d\mu_{L}\circ st_{\hat{d}}^{-1}.$

Since $\mu=\mu_{L}\circ st_{\hat{d}}^{-1}$ and
$\frac{d\nu}{d\mu}=\mathbb{E}_{st_{\hat{d}}^{-1}}st(f^{\omega})\circ
st_{\hat{d}}^{-1}$, this proves the lemma. ∎

######  Remark.

In particular, we will apply Lemmas 32 and 33 in the case where
$\varphi(x)=x\log x-x+1$. This is a convex function from $\mathbb{R}_{+}$ to
$\mathbb{R}_{+}$ which is polynomially bounded, and with this choice of
$\varphi$, we have (assuming that $\nu$ and $\mu$ are probability measures)

$\displaystyle\mathcal{F}(\nu)$
$\displaystyle=\int\left(\frac{d\nu}{d\mu}\log\frac{d\nu}{d\mu}-\frac{d\nu}{d\mu}+1\right)d\mu$
$\displaystyle=\int\frac{d\nu}{d\mu}\log\frac{d\nu}{d\mu}d\mu=H(\nu\mid\mu).$

######  Lemma 34.

(special liftings of bounded densities) Let $\mu\in\mathcal{P}_{2}(\hat{X})$,
and let $\nu\in\mathcal{P}_{2,ac(\mu)}(\hat{X})$ have density
$f=\frac{d\nu}{d\mu}$. Suppose that $f$ is bounded, and $E=supp(f)\subseteq
supp(\mu)$ has bounded diameter. Then, given any
$\mu^{\omega}\in\mathcal{P}_{2}(X)^{\omega}$ such that
$\text{Loeb}(\mu^{\omega})$ pushes forward to $\mu$, there exists an
$S$-integrable lifting $\breve{f}^{\omega}$ of $f\circ
st_{\hat{d}}:st_{\hat{d}}^{-1}(\text{supp}(\nu))\rightarrow\mathbb{R}$ such
that the internal measure

$\nu^{\omega}(A):=\int_{A}\breve{f}^{\omega}d\mu^{\omega}\quad\forall
A\in\mathcal{B}^{\omega}$

is an internal probability measure, belonging to
$\mathcal{P}_{2}(X)_{\lim}^{\omega}$, and such that the $\hat{W}_{2}$-pushdown
of $\nu^{\omega}$ is $\nu$.

###### ?proofname?.

This can be deduced from Lemma 27 together with Lemma 29.

Let $\mu_{L}$ denote $\text{Loeb}(\mu^{\omega})$ and suppose $\mu=\mu_{L}\circ
st_{\hat{d}}^{-1}$. By change of variables, we know that for all Borel
$B\subset supp(\mu)\subseteq\hat{X}$,

$\nu(B)=\int_{B}fd(\mu_{L}\circ
st_{\hat{d}}^{-1})=\int_{st_{\hat{d}}^{-1}(B)}(f\circ st_{\hat{d}})d\mu_{L}.$

From the fact that $f\circ st_{\hat{d}}$ is $\mu_{L}$-integrable and bounded,
we know that there exists a lifting $f^{\omega}$ of $f\circ st_{\hat{d}}$ such
that $f^{\omega}$ is bounded except on a set of $\mu_{L}$-measure zero. This
is not quite enough for us, so we tweak $f^{\omega}$ in several ways.

First, we restrict $f^{\omega}$ to be strictly positive only on a set of
finite diameter. (This is not guaranteed by the fact that $f\circ
st_{\hat{d}}$ is strictly positive only on $st_{\hat{d}}^{-1}(E)$! Away from
this set, it may be the case that $f^{\omega}$ takes nonzero infinitesimal
values.) So let $E^{\omega}$ be an internal set of bounded diameter containing
$st_{\hat{d}}^{-1}(E)$ (such a set exists since $E$, and therefore
$st_{\hat{d}}^{-1}(E)$, has bounded diameter); define

$f^{\omega}\upharpoonright_{E^{\omega}}(x^{\omega})=\begin{cases}f^{\omega}(x^{\omega})&x^{\omega}\in
E^{\omega}\\\ 0&\text{else}.\end{cases}$

We observe that $f^{\omega}\upharpoonright_{E^{\omega}}$ is itself an internal
function, since in particular it has the representation
$f^{\omega}\upharpoonright_{E^{\omega}}=\left[\left(f_{i}\upharpoonright_{E_{i}}\right)\right]$,
where $f^{\omega}=[(f_{i})]$ and $E^{\omega}=[(E_{i})]$. Moreover, since
$f^{\omega}\approx f^{\omega}\upharpoonright_{E^{\omega}}$ pointwise, we know
that $f^{\omega}\upharpoonright_{E^{\omega}}$ is also an $S$-integrable
lifting of $f\circ st_{\hat{d}}$.

Next, let $C=\|f\|_{L^{\infty}(\mu)}$. Define

$\tilde{f}^{\omega}(x)=\begin{cases}f^{\omega}\upharpoonright_{E^{\omega}}(x)&f^{\omega}\upharpoonright_{E^{\omega}}(x)\leq
2C\\\ 2C&\text{otherwise}.\end{cases}$

Note that $\tilde{f}^{\omega}$ is internally defined, by the same reasoning as
for $f^{\omega}\upharpoonright_{E^{\omega}}$. It is also bounded (not even
just bounded $\mu^{\omega}$-a.s.); and obviously differs with $f^{\omega}$
only on a $\mu_{L}$-null set. It is therefore also a lifting of $f\circ
st_{\hat{d}}$. And since $0\leq\tilde{f}^{\omega}\leq f^{\omega}$ pointwise,
and $f^{\omega}$ is $S$-integrable, we know that $\tilde{f}^{\omega}$ is also
$S$-integrable.

Now, it follows, since $\tilde{f}^{\omega}$ is a lifting of $f\circ
st_{\hat{d}}$, that

$\int\tilde{f}^{\omega}d\mu^{\omega}\approx\int f\circ
st_{\hat{d}}d\mu_{L}=1.$

Therefore, we modify $\tilde{f}^{\omega}$ once more, by setting

$\breve{f}^{\omega}:=\frac{1}{\int\tilde{f}^{\omega}d\mu^{\omega}}\tilde{f}^{\omega}.$

With this normalization it is clear that $\breve{f}^{\omega}d\mu^{\omega}$ is
an internal probability measure, and since
$\breve{f}^{\omega}\approx\tilde{f}^{\omega}$ pointwise it is still the case
that $\breve{f}^{\omega}$ is an $S$-integrable lifting of $f\circ
st_{\hat{d}}$.

Finally, let $\nu^{\omega}:=\breve{f}^{\omega}d\mu^{\omega}$. Since
$\nu^{\omega}$ is an internal probability measure with bounded support, it
automatically belongs to $W_{p}(X)_{\lim}^{\omega}$ for every
$p\in[1,\infty)$, in particular for $p=2$. It follows from Lemma 29 (3) that

$\nu_{L}=\int st\circ\breve{f}^{\omega}d\mu_{L}=\int f\circ
st_{\hat{d}}d\mu_{L}.$

Likewise,

$\nu_{L}\circ st_{\hat{d}}^{-1}=\int fd(\mu_{L}\circ st_{\hat{d}}^{-1})=\int
fd\mu=\nu$

so since $\nu^{\omega}$ has bounded support, we know from Lemma 27 that
$\nu^{\omega}$ is automatically a $\hat{W}_{2}$-lifting of $\nu$. Hence, the
proof is complete.

∎

The preceding lemma gets us most of the way to deducing the main theorem of
this section:

###### Proof of Theorem 28.

To reduce notational burden, we suppress reference to the pointed metric
measure isomorphism referred to in Definition 9, and work only with the
ultralimit metric space $(\hat{X},\hat{d})$ equipped with a reference measure
$\mu$. No loss of generality occurs since by [44, Proposition 4.12], the
$CD(K,\infty)$ property is invariant with respect to pmm isomorphism.

Suppose that $\omega$-almost all the $(X_{i},d_{i},\mu_{i})$’s are
$CD(K_{i},\infty)$, and suppose that the Loeb measure $\mu_{L}$ associated to
$\mu^{\omega}=[(\mu_{i})]$ pushes forward to a Radon measure on $\hat{X}$,
that is, $\mu:=\mu_{L}\circ st_{\hat{d}}^{-1}\in\mathcal{P}(\hat{X})$. Assume
that $st(K^{\omega})\neq-\infty$ otherwise there is nothing to prove. Let
$v_{0},v_{1}\in\mathcal{P}_{2,ac(\mu)}(\hat{X})$, with $\mu$-densities $f_{0}$
and $f_{1}$. By [44, Remark 4.6], it suffices to show that

$H(\nu_{\frac{1}{2}}\mid\mu)\leq\frac{1}{2}H(\nu_{0}\mid\mu)+\frac{1}{2}H(\nu_{1}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0},\nu_{1}).$

Step 1. First, let us proceed under the assumption that both $f_{0}$ and
$f_{1}$ are uniformly bounded, and also have compact support. Let
$f_{0}^{\omega}$ and $f_{1}^{\omega}$ be liftings of $f_{0}\circ st_{\hat{d}}$
and $f_{1}\circ st_{\hat{d}}$ respectively, chosen according to Lemma 34.

Now, consider sequences $(f_{0,i})_{i\in\mathbb{N}}$ and
$(f_{1,i})_{i\in\mathbb{N}}$ of functions from $X_{i}$ to $\mathbb{R}_{+}$,
such that $[(f_{0,i})]=f_{0}^{\omega}$ and $[(f_{1,i})]=f_{1}^{\omega}$. Note
that $f_{0,i}d\mu_{i}$ and $f_{1,i}d\mu_{i}$ belong to
$\mathcal{P}_{2,ac(\mu_{i})}(X_{i})$ and have bounded densities for
$\omega$-almost all $i\in\mathbb{N}$. From the fact that $X_{i}$ is
$CD(K_{i},\infty)$, there exists a $W_{2}$-geodesic connecting
$f_{0,i}d\mu_{i}$ and $f_{1,i}d\mu_{i}$ along which the relative entropy
$H(\cdot\mid\mu_{i})$ is $K_{i}$-convex; note that automatically, this
geodesic must have density with respect to $\mu_{i}$ for all $t\in[0,1]$
(otherwise $H(\cdot\mid\mu_{i})$ blows up to $\infty$ at some intermediate
time, contradicting $K_{i}$-convexity). Let $(f_{t,i}d\mu_{i})_{t\in[0,1]}$
denote this distinguished constant speed geodesic; observe that the estimate
of Rajala from [40, Theorem 1.3] shows that for $\omega$-almost all
$i\in\mathbb{N}$,

$\|f_{t,i}\|_{L^{\infty}(\mu_{i})}\leq
e^{K_{i}^{-}D_{i}^{2}/12}(\|f_{0,i}\|_{L^{\infty}(\mu_{i})}+\|f_{1,i}\|_{L^{\infty}(\mu_{i})})$

where $D_{i}=diam(supp(f_{0,i})\cup supp(f_{1,i}))$.

By Łoś’s theorem (in particular Lemma 29 (2)), this implies that
$f_{\frac{1}{2}}^{\omega}:=[(f_{\frac{1}{2},i})]$ is bounded by
$e^{K^{-^{\omega}}(D^{\omega})^{2}/12}(\max f_{0}^{\omega}+\max
f_{1}^{\omega})$, except on a set of $\mu^{\omega}$-measure zero (unless we
are in the degenerate case where $st(K^{\omega})=-\infty$). Note that

$\displaystyle D^{\omega}$ $\displaystyle=diam(supp(f_{0}^{\omega})\cup
supp(f_{1}^{\omega}))<\infty$

since the liftings $f_{0}^{\omega}$ and $f_{1}^{\omega}$ have been selected in
accordance with Lemma 34. It follows that

$st\left(\|f_{\frac{1}{2}}^{\omega}\|_{L^{\infty}(\mu^{\omega})}\right)\leq
st\left(e^{-K^{\omega^{-}}D^{\omega^{2}}/12}\left(\|f_{0}^{\omega}\|_{L^{\infty}(\mu^{\omega})}+\|f_{1}^{\omega}\|_{L^{\infty}(\mu^{\omega})}\right)\right)<\infty.$

This implies that $f_{\frac{1}{2}}^{\omega}$ is $S$-integrable, so that
$st\circ f_{\frac{1}{2}}^{\omega}$ is $\mu_{L}$-integrable. In particular,
Lemma 29 (3) allows us to deduce that

$\forall
B\in\sigma(\mathcal{B}^{\omega})\quad\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})(B)=\int_{B}st\circ
f_{\frac{1}{2}}^{\omega}d\mu_{L}.$

And, from Lemma 31, we know that
$\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ st_{\hat{d}}^{-1}$ is
well-defined (and moreover
$\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1}\ll\mu_{L}\circ st_{\hat{d}}^{-1}=\mu$).

At the same time, we know that

$W_{2}(f_{0,i}d\mu_{i},f_{\frac{1}{2},i}d\mu_{i})=W_{2}(f_{\frac{1}{2},i}d\mu_{i},f_{1,i}d\mu_{i})=\frac{1}{2}W_{2}(f_{0,i}d\mu_{i},f_{1,i}d\mu_{i})\quad\omega\text{-a.s.}$

and hence

$W_{2}^{\omega}(f_{0}^{\omega}d\mu^{\omega},f_{\frac{1}{2}}^{\omega}d\mu^{\omega})=W_{2}^{\omega}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega},f_{1}^{\omega}d\mu^{\omega})=\frac{1}{2}W_{2}^{\omega}(f_{0}^{\omega}d\mu^{\omega},f_{1}^{\omega}d\mu^{\omega}).$

On the one hand, since $f_{0}^{\omega}d\mu^{\omega}$ and
$f_{1}^{\omega}d\mu^{\omega}$ are $\hat{W}_{2}$-liftings of $f_{0}d\mu$ and
$f_{1}d\mu$, we know that

$W_{2}^{\omega}(f_{0}^{\omega}d\mu^{\omega},f_{1}^{\omega}d\mu^{\omega})\approx
W_{2}(f_{0}d\mu,f_{1}d\mu).$

On the other hand, it follows from Propositions 21 and 23 that

$W_{2}(f_{0}d\mu,\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1})\lessapprox
W_{2}^{\omega}(f_{0}^{\omega}d\mu^{\omega},f_{\frac{1}{2}}^{\omega}d\mu^{\omega})$

and

$W_{2}(\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1},f_{1}d\mu)\lessapprox
W_{2}^{\omega}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega},f_{1}^{\omega}d\mu^{\omega}).$

It follows that

$W_{2}(f_{0}d\mu,\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1})\lessapprox\frac{1}{2}W_{2}(f_{0}d\mu,f_{1}d\mu)$

and

$W_{2}(\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1},f_{1}d\mu)\lessapprox\frac{1}{2}W_{2}(f_{0}d\mu,f_{1}d\mu).$

But in these last two displayed equations, all the quantities are real. Hence

$W_{2}(f_{0}d\mu,\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1})\leq\frac{1}{2}W_{2}(f_{0}d\mu,f_{1}d\mu)$

and

$W_{2}(\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1},f_{1}d\mu)\leq\frac{1}{2}W_{2}(f_{0}d\mu,f_{1}d\mu)$

which implies that $\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1}$ must be a $W_{2}$-midpoint between $f_{0}d\mu$ and
$f_{1}d\mu$. So henceforth, we denote
$\nu_{\frac{1}{2}}:=\text{Loeb}(f_{\frac{1}{2}}^{\omega}d\mu^{\omega})\circ
st_{\hat{d}}^{-1}$ and write
$f_{\frac{1}{2}}=\frac{d\nu_{\frac{1}{2}}}{d\mu}$.

Now, returning to the spaces $(\mathcal{P}(X_{i}),W_{2})$, denote
$\nu_{t,i}:=f_{t,i}d\mu_{i}$ for $t=0,\frac{1}{2},1$. Observe that for each
$i$, the $CD(K_{i},\infty)$ property implies that

$H(\nu_{\frac{1}{2},i}\mid\mu_{i})\leq\frac{1}{2}H(v_{0,i}|\mu_{i})+\frac{1}{2}H(v_{1,i}|\mu_{i})-\frac{1}{8}K_{i}W_{2,i}^{2}(\nu_{0,i},\nu_{1,i}).$

From this, we deduce that in the space
$(\mathcal{P}(X)^{\omega},W_{2}^{\omega})$,

$H^{\omega}([(\nu_{\frac{1}{2},i})]\mid[(\mu_{i})])\leq\frac{1}{2}H^{\omega}([(v_{0,i})]\mid[(\mu_{i})])+\frac{1}{2}H^{\omega}([(v_{1,i})]\mid[(\mu_{i})])-\frac{1}{8}K^{\omega}(W_{2}^{\omega})^{2}([(\nu_{0,i})],[(\nu_{1,i})]).$

Note that

$H^{\omega}([(v_{0,i})]\mid[(\mu_{i})])=\int
f_{0}^{\omega}\log^{\omega}f_{0}^{\omega}d\mu^{\omega}\text{ and
}H^{\omega}([(v_{1,i})]\mid[(\mu_{i})])=\int
f_{1}^{\omega}\log^{\omega}f_{1}^{\omega}d\mu^{\omega}$

and so, by Lemma 32, we have that

$H^{\omega}([(v_{0,i})]\mid[(\mu_{i})])\approx H(\nu_{0}\mid\mu)\text{ and
}H^{\omega}([(v_{1,i})]\mid[(\mu_{i})])\approx H(\nu_{1}\mid\mu).$

At the same time, $(W_{2}^{\omega})([(\nu_{0,i})],[(\nu_{1,i})])\approx
W_{2}(\nu_{0},\nu_{1})$ since $[(\nu_{0,i})]$ and $[(\nu_{1,i})]$ are liftings
of $\nu_{0}$ and $\nu_{1}$. Lastly,
$H^{\omega}([(v_{\frac{1}{2},i})]|[(\mu_{i})])\gtrapprox
H(\nu_{\frac{1}{2}}\mid\mu)$, from Lemma 33.

Therefore, since

$H^{\omega}([(\nu_{\frac{1}{2},i})]\mid[(\mu_{i})])\leq\frac{1}{2}H^{\omega}([(v_{0,i})]\mid[(\mu_{i})])+\frac{1}{2}H^{\omega}([(v_{1,i})]\mid[(\mu_{i})])-K^{\omega}\frac{1}{8}(W_{2}^{\omega})^{2}([(\nu_{0,i})],[(\nu_{1,i})])$

and:

  1. (1)

$H^{\omega}([(v_{0,i})]\mid[(\mu_{i})])\approx H(\nu_{0}\mid\mu)$,

  2. (2)

$H^{\omega}([(v_{1,i})]\mid[(\mu_{i})])\approx H(\nu_{1}\mid\mu)$,

  3. (3)

$H^{\omega}([(v_{\frac{1}{2},i})]\mid[(\mu_{i})])\gtrapprox
H(\nu_{\frac{1}{2}}\mid\mu)$,

  4. (4)

$(W_{2}^{\omega})([(\nu_{0,i})],[(\nu_{1,i})])\approx W_{2}(\nu_{0},\nu_{1})$,
and

  5. (5)

$K^{\omega}\approx st(K^{\omega})$,

we deduce that

$H(\nu_{\frac{1}{2}}\mid\mu)\lessapprox\frac{1}{2}H(\nu_{0}\mid\mu)+\frac{1}{2}H(\nu_{1}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0},\nu_{1}).$

But everything in the above expression is a $\mathbb{R}$-valued
function/quantity, so in fact

$H(\nu_{\frac{1}{2}}\mid\mu)\leq\frac{1}{2}H(\nu_{0}\mid\mu)+\frac{1}{2}H(\nu_{1}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0},\nu_{1})$

as desired.

Step 2. The argument we have just given works under the assumption that
$\nu_{0}$ and $\nu_{1}$ have bounded densities and compact support. We now
give an approximation argument that extends the result to general
$\nu_{0},\nu_{1}\in\mathcal{P}_{2,ac(\mu)}(\hat{X})$.

Let $\nu_{0},\nu_{1}\in\mathcal{P}_{2,ac(\mu)}(\hat{X})$ with densities
$f_{0}$ and $f_{1}$. Let $(\kappa_{0,m})_{m\in\mathbb{N}}$ and
$(\kappa_{1,m})$ be sequences of compact sets contained in $supp(\nu_{0})$ and
$supp(\nu_{1})$ respectively, such that

$\bigcup_{m\in\mathbb{N}}\kappa_{0,m}=supp(\nu_{0})\text{ and
}\bigcup_{m\in\mathbb{N}}\kappa_{1,m}=supp(\nu_{1}).$

Define

$(f_{0}\upharpoonright
m)(x):=\begin{cases}\min\\{m,f_{0}(x)\\}&x\in\kappa_{0,m}\\\
0&x\notin\kappa_{0,m}\end{cases}$

and similarly

$(f_{1}\upharpoonright
m)(x):=\begin{cases}\min\\{m,f_{1}(x)\\}&x\in\kappa_{1,m}\\\
0&x\notin\kappa_{1,m}\end{cases}.$

In other words, we both cut off the density at height $m$, and restrict to
some compact set, simultaneously. We also use the notation
$\mu(f_{0}\upharpoonright m):=\int_{\hat{X}}(f_{0}\upharpoonright m)d\mu$ and
similarly for $f_{1}$. Without loss of generality, we assume
$\mu(f_{0}\upharpoonright 1)$ and $\mu(f_{1}\upharpoonright 1)$ are both
strictly positive.

Note that

$H((f_{0}\upharpoonright m)d\mu\mid\mu)=\int_{\hat{X}}(f_{0}\upharpoonright
m)\log(f_{0}\upharpoonright m)d\mu$

still makes sense even though the density $f_{0}\upharpoonright m$ does not
integrate to 1. Clearly $f_{0}\upharpoonright m$ converges pointwise
monotonically to $f_{0}$ as $m\rightarrow\infty$, and similarly for $f_{1}$.
By [5, Lemma 5.1], this implies that

$H((f_{0}\upharpoonright m)d\mu\mid\mu)\rightarrow H(\nu_{0}\mid\mu)\text{ and
}H((f_{1}\upharpoonright m)d\mu\mid\mu)\rightarrow H(\nu_{1}\mid\mu).$

Therefore,

$\frac{1}{\mu(f_{0}\upharpoonright m)}H((f_{0}\upharpoonright
m)d\mu\mid\mu)\rightarrow H(\nu_{0}\mid\mu)\text{ and
}\frac{1}{\mu(f_{1}\upharpoonright m)}H((f_{1}\upharpoonright
m)d\mu\mid\mu)\rightarrow H(\nu_{1}\mid\mu)$

and so

$H\left(\frac{1}{\mu(f_{0}\upharpoonright m)}(f_{0}\upharpoonright
m)d\mu\mid\mu\right)\rightarrow H(\nu_{0}\mid\mu)\text{ and
}H\left(\frac{1}{\mu(f_{1}\upharpoonright m)}(f_{1}\upharpoonright
m)d\mu\mid\mu\right)\rightarrow H(\nu_{1}\mid\mu).$

At the same time, the monotone convergence theorem implies that

$\int_{\hat{X}}f_{0}\upharpoonright md\mu\rightarrow\int f_{0}d\mu\text{ and
}\text{$\int_{\hat{X}}f_{1}\upharpoonright md\mu\rightarrow\int f_{1}d\mu$}$

and hence

$\frac{1}{\mu(f_{0}\upharpoonright m)}\int_{\hat{X}}f_{0}\upharpoonright
md\mu\rightarrow f_{0}d\mu\text{ and }\frac{1}{\mu(f_{1}\upharpoonright
m)}\text{$\int_{\hat{X}}f_{1}\upharpoonright md\mu\rightarrow\int
f_{1}d\mu.$}$

All the terms in the sequences

$\left(\frac{1}{\mu(f_{0}\upharpoonright m)}f_{0}\upharpoonright
m\right)_{m\in\mathbb{N}}\text{ and }\left(\frac{1}{\mu(f_{1}\upharpoonright
m)}f_{1}\upharpoonright m\right)_{m\in\mathbb{N}}$

are dominated by $\frac{1}{\mu(f_{0}\upharpoonright 1)}f_{0}$ and
$\frac{1}{\mu(f_{1}\upharpoonright 1)}f_{1}$ respectively. So by the dominated
convergence theorem,

$\int_{\hat{X}}\left|f_{0}-\frac{1}{\mu(f_{0}\upharpoonright
m)}f_{0}\upharpoonright m\right|d\mu\rightarrow 0\text{ and
}\int_{\hat{X}}\left|f_{1}-\frac{1}{\mu(f_{1}\upharpoonright
m)}f_{1}\upharpoonright m\right|d\mu\rightarrow 0.$

This implies that

$\frac{1}{\mu(f_{0}\upharpoonright m)}f_{0}\upharpoonright
md\mu\rightharpoonup f_{0}d\mu\text{ and }\frac{1}{\mu(f_{1}\upharpoonright
m)}f_{1}\upharpoonright md\mu\rightharpoonup f_{1}d\mu.$

Therefore, to show that $W_{2}(f_{0},\frac{1}{\mu(f_{0}\upharpoonright
m)}f_{0}\upharpoonright m)\rightarrow 0$ and similarly for $f_{1}$, it
suffices to check that we also have convergence of 2nd moments. But this is
clear, since the monotone convergence theorem implies that

$\int_{\hat{X}}d(x_{0},x)^{2}f_{0}\upharpoonright md\mu\rightarrow\int
d(x_{0},x)^{2}f_{0}d\mu\text{ and
}\text{$\int_{\hat{X}}d(x_{0},x)^{2}f_{1}\upharpoonright md\mu\rightarrow\int
d(x_{0},x)^{2}f_{1}d\mu$}$

and hence

$\frac{1}{\mu(f_{0}\upharpoonright
m)}\int_{\hat{X}}d(x_{0},x)^{2}f_{0}\upharpoonright md\mu\rightarrow\int
d(x_{0},x)^{2}f_{0}d\mu$

and

$\text{$\frac{1}{\mu(f_{1}\upharpoonright
m)}$$\int_{\hat{X}}d(x_{0},x)^{2}f_{1}\upharpoonright md\mu\rightarrow\int
d(x_{0},x)^{2}f_{1}d\mu$}.$

Let $\nu_{0,m}:=\frac{1}{\mu(f_{0}\upharpoonright m)}f_{0}\upharpoonright
md\mu$ and $\nu_{1,m}:=\frac{1}{\mu(f_{1}\upharpoonright
m)}f_{1}\upharpoonright md\mu$. By Step 1, we know that

$H(\nu_{\frac{1}{2},m}\mid\mu)\leq\frac{1}{2}H(\nu_{0,m}\mid\mu)+\frac{1}{2}H(\nu_{1,m}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0,m},\nu_{1,m})$

where $\nu_{\frac{1}{2},m}$ is the midpoint between $\nu_{0,m}$ and
$\nu_{1,m}$ constructed in Step 1. At the same time, since
$H(\nu_{0,m}\mid\mu)$, $H(\nu_{1,m}\mid\mu)$, $W_{2}^{2}(\nu_{0,m},\nu_{1,m})$
are all convergent as $m\rightarrow\infty$, it follows, in particular, that

$\displaystyle\sup_{m}H(\nu_{\frac{1}{2},m}\mid\mu)$
$\displaystyle\leq\sup_{m}\left[\frac{1}{2}H(\nu_{0,m}\mid\mu)+\frac{1}{2}H(\nu_{1,m}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0,m},\nu_{1,m})\right]$
$\displaystyle<\infty.$

Therefore, it follows, from from the tightness of sublevel sets of the
relative entropy (see for instance [23, Proposition 4.1]) together with
Prokhorov’s theorem, that we can extract a convergent subsequence of
$(\nu_{\frac{1}{2},m})_{m\in\mathbb{N}}$ (which we do not relabel), whose
limit, which we call $\nu_{\frac{1}{2}}$, is a midpoint of $\nu_{0}$ and
$\nu_{1}$.

Finally, by the lower semicontinuity of $H(\cdot\mid\mu)$,

$\displaystyle H(\nu_{\frac{1}{2}}\mid\mu)$
$\displaystyle\leq\liminf_{m\rightarrow\infty}H(\mu_{\frac{1}{2},m}\mid\mu)$
$\displaystyle\leq\liminf_{m\rightarrow\infty}\left[\frac{1}{2}H(\nu_{0,m}\mid\mu)+\frac{1}{2}H(\nu_{1,m}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0,m},\nu_{1,m})\right]$
$\displaystyle=\frac{1}{2}H(\nu_{0}\mid\mu)+\frac{1}{2}H(\nu_{1}\mid\mu)-\frac{1}{8}st(K^{\omega})W_{2}^{2}(\nu_{0},\nu_{1}).$

∎

## Acknowledgements

Part of this work was conducted with the support of NSF DMS grant 1814991;
additionally, for the part of the work completed while the author was at IHÉS,
the author acknowledges the support of Labex CARMIN. The author thanks Robert
Anderson, David Ross, Dejan Slepčev, and Henry Towsner for helpful
discussions. Additionally, the author thanks Timo Schultz for pointing out an
error in a previous version of this manuscript.

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# Origin of the gossamer ferromagnetism in MnTe

I. I. Mazin  Department of Physics and Astronomy, George Mason University,
Fairfax, VA 22030, USA  Quantum Science and Engineering Center, George Mason
University, Fairfax, VA 22030, USA

###### Abstract

Absence of net magnetization in altermagnetis is both a blessing (no stray
fields) and a curse (no obvious way to manipulate altermagnetic domains by
external fields). Yet, MnTe was demonstrated experimentally to have no
measurable stray fields and yet controllable by external magnetic filed — a
win-win situation. In this paper we unravel the microscopic mechanism of this
proprety. It appear to emerge from concerted action of three different
mechanisms. Microscopic understanding of the beneficial properties of MnTe
opens a road to controllable design of similar altermagnets for spintronics
applications.

##  I Introduction

Altermagnetism has attracted a lot of attention recently[1, 2, 3]. Both novel
physical phenomena and potential applications with orders of magnitude better
performance have been discussed. Already more than a hundred of altermagnets
have been identifies among existing antiferromagnets or predicted
theoretically.

Nevertheless, nearly all experimental studies so far have been performed on a
handful of materials. Till recently, the most popular was RuO2, where in 2017
weak antiferromagnetism (Ru moments $\sim 0.05\ \mu_{B}$) was reported in
neutron scattering experiment[4], with an altermagnetic ordering pattern.
Close to 25 papers have been published in the last few years claiming various
altermagnetic manifestations. Unfortunately, in most cases an interpretation
in terms of altermagnetism required and order-of-magnitude moment than that
reported in Ref. [4], and, to add insult to injury, it was recently shown that
even the reported moment was an artifact resulting from multiple neutron
scattering and in reality RuO2 is nonmagnetic both in bulk and in films[5, 6],
albeit may be magnetic in a few atomic layer form[7].

The second in popularity altermagnetic candidate is Mn5Si3. There the problem
is different. The materials is well studied experimentally, and three ordered
magnetic phases exist in the bulk[8, 9, 10], one collinear and two
noncollinear. However, all three order with the vector {0,1/2,0}, so by
definition not altermagnetic. An argument was put forward that in thin films
the order can switch spontaneously to {0,1/2,0}, but the fact that three
different magnetic patterns share the same ordering vector suggests existence
of strong antiferromagnetic coupling between the neighboring along the $b$
direction unit cells, which would be difficult to overcome in thin films.

This leaves with the third by popularity compound, MnTe. This is truly a
poster child for the altermagnetic case. Its magnetic structure is perfectly
well know, and is definitely antiferromagnetic. It is self doped (hole
concentration $\sim 10^{18}$ cm-3[11]), and the Néel vector is in the 210
direction, which is the one compatible with the anomalous transport. As I will
discuss in this paper, it also possesses a fortuitous combination of other
parameters, making it truly an ideal candidate for both studying the
altermagnetic physics and for spintronics applications.

##  II MnTe: Background

Table 1: Four weak effect combining to provide altermagetic domain control by an external field. Here $H$ is the field, $x$ is the hole concentration, $t$ characterizes one-electron hopping amplitude near the Fermi level, $J$ is a gauge of the average Heisenberg coupling, and $M$ is the Mn magnetic moment, $\approx 5\ \mu_{B}$. effect | provides | order of magnitude
---|---|---
Double exchange | weak $z$ component of magnetization | $Mtx/J$
Feedback of spin polarization on charge distribution | symmetry lowering | SOC2
DMI in the symmetry-lowered structure | coupling between ferromagnetism and altermagnetism | SOC
External field | Zeeman coupling with weak ferromagnetism | H

MnTe, despite being a canonical Mott (or, more correctly, charge transfer)
insulator has been known to occur always in a self-doped form, exhibiting a
sizeable conductivity and also showing anomalous Hall conductivity (AHC),
discovered as early as in 1965[12]. The fact that AHC is present in an
antiferromagnetic materials was either ignored[13] or ascribed to weak
ferromagnetism of unknown origin[12]. Recently, the interest to anomalous
transport in MnTe was rekindled after researchers realized that it is
altermagnetic[1, 2, 3, 14]. AHC has been remeasured[15, 16, 17, 18], and weak
ferromagnetism detected and its amplitude estimated to be $\sim 2.5\times
10^{-5}$[18] to $5\times 10^{-5}$[16] $\mu_{B}$/Mn, parallel to $c$.
Furthermore, it has been now realized that the magnetization is too small to
explain the large observed AHC, which then must have the altermagnetic origin.

Regarding the origin of this extremely weak (“gossamer”) ferromagnetism,
different hypotheses were offered. In the earliest paper[12] it was pointed
out that the magnetic space group (MSG) Cm’c’m, which is realized in MnTe[19],
is, per Turov’s classification[20], compatible with ferromagnetism, while no
assumptions were made about the microscopic nature of the latter. In several
later papers[21, 17] it was suggested that excess of Mn may lead to weak
ferromagnetism. Similarly, in Ref. [16] an empirical energy expression,
compatible with the MSG, was introduced, with a remark (quite correct) that
the standard Dzyaloshinskii-Moriya interaction (DMI) cannot generate this
expression.

However, a possibility of excess (interstitial) Mn or Te vacancies seems
unlikely given that in such a case one would expect electron, not hole doping.
So the question of microscopic origin of the weak ferromagnetism remains open.

In this paper, we will demonstrate that concerted action of four separate
mechanisms leads to a small magnetic canting. In a nutshell, there is one
effect that generates the canting, namely double exchange due to hole doping.
Using DFT calculations we can estimate the magnitude of the effect and it
appears to be consistent with the experiment. However, it does not couple to
the altermagnetic order parameter and cannot be used as a “handle” to
manipulate altermagnetic domains.

The second effect (quadratic in spin-orbit coupling) is the feedback of
altermagnetic ordering on the charge distribution, which marginally lowers the
(charge) space group from # 194 (P63/mmc) to #63 (Cmcm).

The third effect is a higher-order DMI interaction that couples the already
existing weak ferromagnetic component with altermagnetism, thus allowing to
manipulate altermagnetic domains with a external magnetic field. This effect
only appears because of the symmetry lowering described above. And, of course,
the fourth effect is rather trivial: Zeeman coupling of the external field
with the ferromagnetic component. This is summarized in Table 1.

##  III Anisotropic exchange coupling in MnTe

Figure 1: Blue: Mn up; purple: Mn down; orange, olive: Te

In this Section, I will construct the full $3\times 3$ exchange matrices for
the first three Mn-Mn bonds, and will show that they cannot induce a uniform
canting. A full exchange matrix for a bond $I=1,2,3$ is defined such that its
energy is

$E^{(I)}=\sum_{\alpha\beta}\mathbf{m_{1}}\cdot\mathcal{J}\cdot\mathbf{m_{2}}^{\prime},$
(1)

where $\mathbf{m}$s are unit vectors in the direction of the Mn moments at the
ends of the bond, and

$\mathcal{J}=\begin{pmatrix}J_{xx}&J_{xy}&J_{xz}\\\ J_{yx}&J_{yy}&J_{yz}\\\
J_{zx}&J_{zy}&J_{zz}\end{pmatrix}$ (2)

For a reference, I quote the calculated values of the isotropic (Heisenberg)
exchange $(J=J_{xx}+J_{yy}+J_{zz})/3$ in these three cases. The nearest
neighbor bond, which is perpendicular to the basal plane, is strongly
antiferromagnetic (AF), $J_{1}=42.1$ meV, the next nearest neighbor bond (the
nearest inside the plane) has $J_{2}\approx 0$, and the third, which is just
the sum of the other two, is also AF, $J_{3}=5.3$ meV.

To begin with, let us show that there is no DMI interaction on the nn (AF)
bond. Indeed, there is a 001 mirror plane through the midpoint, which
indicates that the DMI vector $\mathbf{D}\perp\mathbf{\hat{z}}.$ The are also
three mirrors passing through the bond itself, nalely 100, 010 an 110. D must
be perpendicular to each of them, which is impossible. Moreover, a detailed
analysis shows that the full exchange matrix for this bonds is exceedingly
simple:

$\mathcal{J}=\begin{pmatrix}J&0&0\\\ 0&J&0\\\ 0&0&J+J_{z}\end{pmatrix}$ (3)

i.e., only the $J_{z}$ Ising exchange is allowed besides the Heisenberg
exchange.

Figure 2: Labeling of the Mn atoms (the upper 6) and DMI bonds (bond centers
depicted with grey balls). The same color bonds (and their inversion partners
are equivalent in the local coordinate sistem explained in the text. For
instance, for the bond 1 the local system coincides with the global one,
$x,y,z$, and for the others it is rotated in steps of $\pi/3$

The nn in the planes are not DMI because of inversion. Second nn across the
plane (depicted in Fig. 2) have a $C_{2}$ passing through the midpoint, namely
for the 10$\frac{1}{2}$ bond the axis is 010. D is then perpendicular to this
axis, that is, lies in the $xz$ plane. Let us consider this interaction more
closely. There are 12 bonds of this kind, which I will label as $i,s,$ where
$i=1..6$ are rotations around 001 and $s=\pm$ indicates where the bond is
going up or down. Let me also write the full exchange matrix as

$\hat{J}^{is}=\begin{pmatrix}J_{XX}^{is}&J_{XY}^{is}&J_{Xz}^{is}\\\
J_{YX}^{is}&J_{YY}^{is}&J_{Yz}^{is}\\\
J_{zX}^{is}&J_{zY}^{is}&J_{zz}^{is}\end{pmatrix}$ (4)

where the new axis $X$ is the projection of the bond onto the $ab$ plane, and
$Y$ is perpendicilar to $X,z.$ Note that is is enough to consider only $s=+,$
because due to the inversion $\hat{J}^{1-}=\hat{J}^{4+},$
$\hat{J}^{2-}=\hat{J}^{5+},$ $\hat{J}^{3-}=\hat{J}^{6+},$ etc. As mentioned,
there is only one operation that keeps a bond in place, $C_{2Y},$ which
transforms $S_{X}\rightarrow-S_{X,}$ $S_{Y}\rightarrow S_{Y,}$
$S_{Z}\rightarrow-S_{Z,}$ so

$\displaystyle\begin{pmatrix}J_{XX}&J_{XY}&J_{Xz}\\\ J_{YX}&J_{YY}&J_{Yz}\\\
J_{zX}&J_{zY}&J_{zz}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}J_{XX}&-J_{YX}&J_{zX}\\\
-J_{XY}&J_{YY}&-J_{zY}\\\ J_{Xz}&-J_{Yz}&J_{zz}\end{pmatrix}$ (5)
$\displaystyle=\begin{pmatrix}J_{XX}&D_{z}&A\\\ -D_{z}&J_{YY}&D_{X}\\\
A&-D_{X}&J_{zz}\end{pmatrix},$ (6)

therefore $J_{Xz}=J_{zX}=A,$ $J_{XY}=-J_{YX}=D_{z},$ $J_{zY}=-J_{zX}=D_{X}.$

Next, consider a mirror plane through the origin and the Te atoms. One such
plane converts $J^{1}$ to $J^{2},$ and it also changes $S_{X}\rightarrow-
S_{X,}$ $S_{Y}\rightarrow S_{Y,}$ $S_{Z}\rightarrow-S_{Z}.$ Then

$\begin{pmatrix}J_{XX}^{1}&D_{z}^{1}&A^{1}\\\
-D_{z}^{1}&J_{YY}^{1}&D_{X}^{1}\\\
A^{1}&-D_{X}^{1}&J_{zz}^{1}\end{pmatrix}=\begin{pmatrix}J_{XX}^{2}&-D_{z}^{2}&A^{2}\\\
D_{z}^{2}&J_{YY}^{2}&-D_{X}^{2}\\\ A^{2}&D_{X}^{2}&J_{zz}^{2}\end{pmatrix},$
(7)

That is, the compass term $A$ is the same, the $z$-component of the DMI vector
changes sign, and also the radial vector. The result is shown in Fig. 3, where
the red arrows give the local $X$ axis (the radial component of the DMI
vector, blue, is parallel). Since the altermagnetic Néel vector lies in the
plane, the compass terms has no effect, and the DMI field exhorted by the
parallel to each other spin-up moments onto the the spin down subsystem cancel
out. Note that this cancellation is a direct consequence of the 3-fold
rotational axis.

Figure 3: DMI interaction. Dotted and crossed corcles show $\pm D_{z}$, the
blue arrows $D_{X}$, and the dashed black ones the small (tiny!) components
that arise from the feedback of the magnetic symmetry breaking. Figure 4:
Difference between the charge densities of the 210 Néel vector (the ground
state) and the 100 one. The colors indicate the sign, and isosurfaces are
drawn at the level of $2\times 10^{-6}$e/Å3 (about 4 orders of magnitude less
than the total charge density). Note breaking of the $C_{3}$ rotational
symmetry.

However, that does not account for the magnetic order already affecting the
crystal symmetry. As long as there is spin-orbit coupling, there will be an
additional very small change in the charge distribution ($\propto M^{2})$ that
breaks the hexagonal symmetry of the charge cloud, as shown (exaggerated) in
Fig. 4.

In Fig. 4 I show results of DFT calculation using the VASP code[22, 23].
Charge density was calculated self-consistently, including SOC, for the Néel
vector $\mathbf{M}||210$ and the perpendicular to it vector $\mathbf{M}||100$.
A very fine $k$-point mesh of $21\times 21\times 11$ was used, with the
convergence criterion of $10^{-8}$ eV and an energy cut-off of 800 eV. In
order to avoid systematic error, both calculations were started with a self-
consistent calculation with $\mathbf{M}||001$, and then the spin coordinate
system was rotated using the SAXIS tag in VASP. The same protocol applied to
nonrelativistic calculations did not produce any symmetry breaking.

Because of this symmetry breaking, the mirror $\sigma_{Y}$ remains only for
the bonds $J^{1}$ and $J^{4},$ but is broken for the others (and, of course,
$C_{3}$ is broken as well). As a result, the in-plane vector $D_{x}$ is not
strictly radial any more for the other bonds, but acquires a tiny tangential
component $D_{y}.$

Let me now concentrate on the bonds $J^{2}$ and $J^{3}.$There still is a
mirror relating the two bonds. However, there is no condition any more that
$D_{Y}=0.$ Keeping only the DMI terms,

$\begin{pmatrix}&D_{z}^{3}&D_{Y}^{3}\\\ -D_{z}^{3}&&D_{X}^{3}\\\
-D_{Y}^{3}&-D_{X}^{3}&\end{pmatrix}=\begin{pmatrix}&-D_{z}^{2}&D_{Y}^{2}\\\
D_{z}^{2}&&-D_{X}^{2}\\\ -D_{Y}^{2}&D_{X}^{2}&\end{pmatrix},$ (8)

That is to say, there appears a small tangential component $D_{Y}.$

Now we need to connect that to $J^{5,6}.$ There is a $C_{2x}$ axis. It will
convert $J^{2}$ to $J^{6},$ and, in the original Cartesian system,
$S_{x}\rightarrow S_{x,}$ $S_{y}\rightarrow-S_{y,}$ $S_{z}\rightarrow-S_{z}.$
Without going into details, it means that the $J_{yz}$ component will not
change, that is, $D_{x}$ will remain the same, and will have a net component
after averaging, as shown in Fig. 3 by the dashed arrows.

While we do not make any attempt to cast this interaction into a gauge-
invariant form, from general symmetry considerations one can conclude that
this interaction can be written generically as
$H=\sum_{ij}{(\mathbf{S}_{i}\cdot\hat{K}\cdot\mathbf{S}_{i})(\mathbf{S}_{i}\times\mathbf{S}_{j})}\cdot\mathbf{C}_{ij}$,
where the tensor $\hat{K}$ is similar to the single-site anisotropy tensor,
and $\mathbf{C}_{ij}$ similar to the Dzyaloshinskii-Moriya exchange. It can be
compared to the interaction defined by Blügel et al[24] and by Brinker et
al[25] (see also Ref. [26]) as “chiral biquadratic interaction”,
$H=\sum_{ij}{(\mathbf{S}_{i}\cdot\mathbf{S}_{j})(\mathbf{S}_{i}\times\mathbf{S}_{j})}\cdot\mathbf{C}_{ij}$.

##  IV Double exchange

Double exchange[27] is a competition between antiferromagnetic exchange among
(reasonably well) localized moments, which wants to keep the latter
antiparallel, and the kinetic energy of the itinerant electrons, assuming that
there is a strong interaction keeping itinerant electrons’ spins locally
parallel to the localized moments. This interaction can be Hund’s rule
coupling, or, if the itinerant electrons originate from different atomic
species, Schrieffer-Wolff interaction[28]. The corresponding Hamiltonian can
be written as[27]:

$E=-aJ\cos{2\phi}-bxt\sin{\phi},$ (9)

where $E$ is the total energy per spin, $J$ is a properly averaged
antiferromagnetic exchange coupling, $x$ is the concentration of itinerant
carriers (assumed $x\ll 1$), $t$ is the hopping parameter of the itinerant
electrons, and $a$ and $b$ are geometrical coefficients characterizing the
connectivity in the system. $\phi$ is the canting angle away from the original
collinear antiferromagnetic arrangement. Minimizing this expression, one gets

$\sin{\phi}=(bt/4aJ)x,$ (10)

that is to say, the net ferromagnetic moment is linearly proportional to the
concentration $x$.

In order to access this effect from DFT, we have performed a series of
calculations fixing the electron count at $-x$ compared to the stoichiometric
compound, and computed the energy difference between the parallel and
antiparallel configurations, i.e., $\phi=0$ or $\pi/2$ (all computational
parameters were the same as in Ref. [14]). Per Eq. 5, it is

$\Delta E=2aJ-bxt.$ (11)

Fig. 5 shows the result. Indeed, $\Delta E$ is perfectly linear with $x$ (and
only slightly deviates from linearity even at $x=0.25)$. From the slope of the
line in Fig. 5 we deduce that $(bt/4aJ)=2.8$, and the magnetic moment for the
hole concentration $\sim 10^{18}$ cm-3 (per Ref. [11]) the weak FM moment is
$m\sim(btx/4aJ)\times 5=7\times 10^{-4}\ \mu_{B},$ somewhat larger, but in a
very reasonable agreement with experiment (particularly taking into account
that the concentration and magnetization were measured by different group on
different samples).

Figure 5: Energy difference between the FM and AF states, per Mn, as described
in Eq. 11.

##  V Discussion and conclusions

A major obstacle on the road to exciting spintronics applications is inability
to manipulate altermagnetic domains by external stimuli. Indeed, the major
advantage of altermagnets over ferromagnets is absence of net magnetization
and thus of stray fields. Yet the same zero net magnetization prevents an
altermagnet from interacting with external magnetic fields and thus presents a
problem with controlling and manipulating altermagnetic domains. In many
cases, in fact in those same cases where anomalous transport in an altermagnet
is nonzero, the magnetic point group is compatible with ferromagnetism, a
phenomenon know as weak ferromagnetism. However, several prerequisites are
necessary before one can take an advantage of this weak ferromagnetic
component, namely, the canting of altermagnetic moments need not only be
allowed by symmetry, but microscopic interactions must be present generating
such canting. Second, this canting must be small enough not to generate stray
fields by itself. Third, it has to couple with the altermagnetic order so that
flipping the ferromagnetic component would lead to flipping the altermagetic
order.

Three mechanisms are known to generate weak ferromagnetism. One is a single-
site anisotropy, as in NiF2[29], another double exchange. Neither couples with
the altermagnetic order, so is useless. One possibility is Dzyaloshinskii-
Moriya interaction, but, as the discussion above exemplifies, absence of an
inversion center at the midpoint of a magnetic bond does not guarantee
(contrary to a common misconception) presence of such interaction, and even if
it is non-zero in most cases it cancels out when summed over all equivalent
bonds. Besides, if it is allowed, it is the first order in spin-orbit coupling
so may actually be strong enough to generate too large stray fields.

As discussed in this paper, MnTe represents a new paradigm, applicable to
altermagnetic semiconductors (actually the largest subclass of the latter),
doped naturally or in a controlled way by a very small number of carriers,
$10^{-4}$–$10^{-5}$ per magnetic ion. This doping plays a double (positive)
role: first, it ensures metallic conductivity required for applications.
Second, it generates a very small ferromagnetic canting in a controllable way.
Thus, we have three decoupled, that is, controlled by different factors,
parameters: altermagnetic transport, which is completely oblivious to the
small ferromagnetic component, weak ferromagnetism controlled by the carrier
concentration, and the four-spin Dzyaloshinskii-Moriya (“chiral biquadratic”)
interaction that couples the ferro- and and the altermagnetic order
parameters. The fact that MnTe really exists and experimentally demonstrates
all these features is strongly encouraging and suggest a new road towards
altermagnetic spintronics.

###### Acknowledgements.

This work was supported by the Army Research Office under Cooperative
Agreement Number W911NF-22-2-0173. The author is thankful to Nirmal Ghimire,
Libor Ŝmejkal, Jairo Sinova, Kirill Belashchenko and Stefan Blügel for
valuable discussions.

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