Notation in Huybrechts' $A\wedge A$

Question

When introducing curvature (p.183), Huybrechts uses a notation which I don't really understand: For example in the formulas $$F_{\nabla}=d(A)+A\wedge A, $$ $$dF_\nabla = F_\nabla \wedge A - A\wedge F_\nabla.$$ What does "$\wedge$" stand for? It doesn't seem to be the usual exterior product, because from the way the formulas are written, "$\wedge$" doesn't seem to be skew-symmetric.

Answer 1

In addition to @Andrew's explanation, $\wedge$'s therein are not only exterior products, but also respective algebraic arithmetic. Here are two examples.

Suppose $A$ is a usual $1$-form, e.g., $A=\omega$. Then $A\wedge A=\omega\wedge\omega=0$ follows the usual exterior product.

Suppose $A$ is a matrix $1$-form, e.g., $$ A=\left( \begin{array}{cc} \omega_{11}&\omega_{12}\\ \omega_{21}&\omega_{22} \end{array} \right), $$ where each $\omega_{ij}$ is a usual $1$-form. Then $$ A\wedge A=\left( \begin{array}{cc} \omega_{11}&\omega_{12}\\ \omega_{21}&\omega_{22} \end{array} \right)\wedge\left( \begin{array}{cc} \omega_{11}&\omega_{12}\\ \omega_{21}&\omega_{22} \end{array} \right)=\left( \begin{array}{cc} \omega_{11}\wedge\omega_{11}+\omega_{12}\wedge\omega_{21}&\omega_{11}\wedge\omega_{12}+\omega_{12}\wedge\omega_{22}\\ \omega_{21}\wedge\omega_{11}+\omega_{22}\wedge\omega_{21}&\omega_{21}\wedge\omega_{12}+\omega_{22}\wedge\omega_{22}\end{array} \right), $$ where $\wedge$ not only includes the usual exterior product, but also the matrix-matrix multiplication.

Answer 2

The wedge here indicates the combination of wedge product on forms and composition of the endomorphisms. So, if we have $A= \sum_i \phi_i\otimes a_i$ and $B=\sum_j \psi_j \otimes b_j$, where $\phi_i,\psi_j$ are forms and $a_i,b_j$ are endomorphisms, then $$ A\wedge B = \sum_{i,j} (\psi_i\wedge \phi_j) \otimes (a_i\circ b_j) $$

Huybrechts mentions this in the middle of page 183, but it didn't fill in this detail.