Correct curvature tensor of symmetric space of positive definite matrices with trace metric? Let $Pos(n)$ be the set of $n \times n$ real positive definite matrices with trace (aka affine-invariant) metric
$$\langle u, v \rangle_p = tr(p^{-1} u p^{-1} v)$$
for all $p \in Pos(n)$ and $u, v \in T_p Pos(n) \cong Sym(n)$.
Here $Sym(n)$ is the set of real symmetric $n \times n$ matrices, which we identify with the tangent spaces of $Pos(n)$.
In Chapter XII of Fundamentals of Riemannian Geometry, Serge Lang states that for $u, v, w \in T_I Pos(n) \cong Sym(n)$, the curvature tensor is given by
$$\langle R(v, w)v, w \rangle = -2 tr((v w)^2-v^2 w^2).$$
However, in papers online (see this and this) Dolcetti et al. show that
$$\langle R(v, w)v, w \rangle = \frac{1}{4}(-2 tr((v w)^2-v^2 w^2)).$$
Note the factor of $\frac{1}{4}$ here.
Which expression for the curvature tensor is correct?

More context:
Dolcetti et al. compute the curvature tensor with fairly direct methods.  Lang uses the following approach.  There is one step of this approach I am uncertain of.
Let $M$ be a symmetric space with Riemannian connection $\nabla$.  Let $p \in M$ and define $m_p$ to be the set of Killing vector fields $X$ on $M$ with $(\nabla X)_p = 0$.
It is well-known that the map
$$m_p \rightarrow T_p M,  X \mapsto X_p$$
is a bijection.
Moreover, it is also well-known that the curvature tensor of $M$ is given by
$$R(U, V) W = [W, [U, V]].$$
for all $U, V, W \in m_p$.
Here $[\cdot, \cdot]$ denotes the Lie bracket.
For references, see
(1) Chapter XIII of Serge Lang's Fundamentals of Riemannian Geometry
or (2) Chapter 8 of Peter Petersen's Riemannian Geometry.
Using these facts, in the last theorem of Chapter XII, Lang gives a formula for curvature tensor of the space of positive definite matrices $Pos$ with the trace metric:

Here Lang seems to implicitly use that for $U, V, W \in m_p$,
$$[W, [U, V]] = [W_p, [U_p, V_p]_{com}]_{com}$$
where for matrices $A, B$ we have $[A, B]_{com} = A B - B A$ is the matrix commutator (not the Lie bracket).
It is not at all clear to me that this is true.
For example, if $U, V \in m_p$, it cannot be the case that $[U, V]_p = U_p V_p - V_p U_p$ because $U_p V_p - V_p U_p$ is not element of $T_p Pos(n) = Sym(n)$.
 A: I am not 100% certain with this, but I think the following computations check out.
Let $x$ be an arbitrary symmetric matrix, then the vector field $X$ that takes value $\frac12(xM + Mx)$ at point $M\in \mathrm{Pos}(n)$ is Killing and satisfies $\nabla X |_{e} = 0$. (That it is Killing follows from the fact that it generates the one parameter family of isometries $\mathbb{R}\times \mathrm{Pos}(n) \ni (t,M) \mapsto \exp(\frac12 t x) M \exp(\frac12 tx)$. That it has vanishing covariant derivative at the identity can be checked using a local coordinate computation.)
Now let $X,Y,Z$ be the vector fields corresponding to $\frac12 (xM + Mx), \frac12 (yM + My), \frac12 (zM + Mz)$ respectively.
The Lie bracket
$$ [X,Y] = \frac14 (xyM + Myx - yxM - Mxy) $$
of course vanishes at the origin, and we find
$$ [[X,Y],Z] |_e = \frac14 ( xyz + zyx - yxz - zxy) = \frac14 ( [xy,z]_{\mathrm{com}} - [yx,z]_{\mathrm{com}}) = \frac14 [[x,y]_{\mathrm{com}},z]_{\mathrm{com}} $$
which would indicate that Lang omitted a factor of 1/4 when converting from the Lie bracket of vector fields to the representation using matrix commutators.
