Positivity of Currents I already asked about this a couple of weeks ago but had introduced some rather annoying notation. I decided to reformulate the question in a more compact format. (edit: old post taken down as it is basically a duplicate now)
Suppose $\psi$ is the complex differential $(1,1)$-form $$\psi = i \sum_{j,k=1}^n h_{j,k} ~dz_j \wedge d\bar{z}_k$$
and that the matrix $H$ with entries $h_{j,k}$ is hermitian (i.e. $\psi$ is a real form). In a lecture of mine, it was given as an exercise that $\psi$ is positive (i.e. $\psi(v,Jv) > 0$ for all $v \ne 0$, where $J$ is the canonical almost complex structure) if and only if $H$ is positive-definite.
This differential form induces a current by integration, $$T_{\psi} ~\colon ~ \Omega^{n-1,n-1}_c(\mathbb{C^n}) \rightarrow \mathbb{C}, ~ \phi \mapsto \int_{\mathbb{C^n}} \psi \wedge \phi.$$
In Harris and Griffiths' textbook "Principles of Algebraic Geometry" (on page 386 with $p=1$), a real $(1,1)$-current is defined to be positive if for every $(n-1,0)$-form $\eta$ we have that $T(\eta \wedge \bar{\eta})$ is a non-negative real number.
Take such a test form $$\eta = \sum_{j=1}^n \phi_j ~dz_J,$$ where $dz_J$ denotes $dz_1 \wedge \dots \wedge \hat{dz_j} \wedge \dots \wedge dz_n$. Then we have $$\eta \wedge \bar{\eta} = \sum_{j,k} \phi_j \overline{\phi_k} ~dz_J \wedge d\bar{z}_K$$ and
\begin{align*}
T_{\psi}(\eta \wedge \bar{\eta}) &= i \sum_{j,k} \int h_{j,k} \phi_j \overline{\phi_k} ~dz_j \wedge d\bar{z}_k \wedge dz_J \wedge d\bar{z}_K \\
&= i \sum_{j,k} \int h_{j,k} \phi_j \overline{\phi_k} ~\sigma(j,k) ~dz_1 \wedge d\bar{z}_1 \wedge \dots \wedge dz_n \wedge d\bar{z}_n,
\end{align*}
where $\sigma(j,k)$ denotes the sign coming from $$dz_j \wedge d\bar{z}_k \wedge dz_J \wedge d\bar{z}_K = \sigma(j,k) ~dz_1 \wedge d\bar{z_1} \wedge \dots \wedge dz_n \wedge d\bar{z}_n.$$
Thus, the current $T_{\psi}$ is positive not if $H$ is positive definite but if the matrix with entries $\sigma(j,k) h_{j,k}$ is positive definite.
Is it true that positivity of currents and positivity of differential forms do not give rise to the same notion? Can this be fixed?
 A: Revisiting my question after a month, I finally realized that the answer is frustratingly simple...
The sign $\sigma(j,k)$ can be written explicitly as
$$ \sigma(j,k) = (-1)^{n-1 + \frac{n(n-1)}{2}} (-1)^{j-1} (-1)^{k-1}. $$
The crucial (yet simple) observation is that the matrix with entries $(-1)^{j+k}h_{j,k}$ is positive definite if and only if the matrix with entries $h_{j,k}$ is.
Thus, the two notions of positivity agree up to a sign that depends only on $n$.
In the following I would like to propose a correction to the definition in Harris and Griffiths' book.
(Remark: to some extend, their definition already seems to be flawed as the exponent of the $i$ does not depend on $n$.)
Let us say that a real $(p,p)$-current $T$ is positive if for any compactly supported $(n-p,0)$-test form $\eta$ we have
$$ (-1)^{\frac{(n-p)(n-p-1)}{2}} i^{n-p} T(\eta \wedge \bar{\eta}) \geq 0. $$
For $T=T_{\psi}$ as in the question we get ($p=1$)
$$ (-1)^{\frac{(n-1)(n-2)}{2}} i^{n-1} T_{\psi}(\eta \wedge \bar{\eta}) =\\
 = i^n \int_{\mathbb{C}^n} \left( \sum_{j,k=1}^n (-1)^{j+k} \phi_j h_{j,k} \bar{\phi}_k \right) dz_1 \wedge d\bar{z}_1 \wedge \dots \wedge dz_n \wedge d\bar{z}_n \\
 = 2^n \int_{\mathbb{C}^n} \left( \sum_{j,k=1}^n (-1)^{j+k} \phi_j h_{j,k} \bar{\phi}_k \right) dx_1 \wedge dy_1 \wedge \dots \wedge dx_n \wedge dy_n, $$
where the sign from the definition of positivity cancels with the sign in $\sigma(j,k)$.
Thus, $T_{\psi}$ is positive if and only if this integral is always positive if and only if $((-1)^{j+k}h_{j,k})$ is positive definite if and only if $(h_{j,k})$ is positive definite if and only if $\psi$ is positive, as desired.
To further motivate the particular choice of sign in the definition (for now, we only verified it for $p=1$), let us consider the current defined by integration over $\mathbb{C}^{n-p} \times \{0\} \subset \mathbb{C}^n$, which ought to be a positive current.
Indeed,
$$ \int_{\mathbb{C}^{n-p} \times \{0\}} \phi \bar{\phi} ~dz_1 \wedge \dots \wedge dz_{n-p} \wedge d\bar{z}_1 \wedge \dots \wedge d\bar{z}_{n-p} \\
 = (-1)^{\frac{(n-p)(n-p-1)}{2}} \int_{\mathbb{C}^{n-p} \times \{0\}} |\phi|^2 ~dz_1 \wedge d\bar{z}_1 \wedge \dots \wedge dz_{n-p} \wedge d\bar{z}_{n-p} \\
 = (-1)^{\frac{(n-p)(n-p-1)}{2}} (-2i)^{n-p} \int_{\mathbb{C}^{n-p} \times \{0\}} |\phi|^2 ~dx_1 \wedge dy_1 \wedge \dots \wedge dx_{n-p} \wedge dy_{n-p}. $$
