Why the cross product between 2 vectors exists only in 3 dimensions? Why the cross product between 2 (two!) vectors exists only in 3 dimensions?
The article "Cross Products of Vectors in Higher Dimensional Euclidean Spaces" by W. M. Massey (see http://doi.org/10.2307/2323537) may be related related to my question, but I have no access to it.
Another reference is posted here: Finding a fourth vector that makes a set a basis, but is about the product of $n-1$ vectors of dimension $n$. In it, the book "Vector Calculus" by Susan J. Colley is mentioned. The answer, however do not mention the specific following phrase of Colley: "There is no simple generalization of the cross product".
I heard in a lecture (without further details, unfortunately) that the answer to my question is because $n=3$ is the only non-negative integer number such that $n = \displaystyle \frac{(n-1)n}{2}$. Why is that? Where does it come from?
By "cross product" I expect to have ($a, b$ and $c$ denoting any $n-$dimensional vectors):


*

*If $c = a \times b$ then $c \cdot a = c \cdot b = 0$, where " $\cdot$ " denotes the usual dot product;

*$a \times b = - \,b \times a$;

*$\| a \times b \|$ is equal to the hipervolume determined by $a$ and $b$.


As mentioned by steven gregory, an orientation is missing.
 A: Formally, the cross product of two vectors in $3$ space should be viewed an an antisymmetric second rank tensor.  In $3$ space these things are also called axial vectors.  The point is that if you invert the space normal vectors change sign, but axial vectors do not.  This is because the cross product is the product of two vectors.  If you change the sign of each, the sign of the product does not change.  Viewing the cross product as a vector works (if we ignore the sign change problem) because an antisymmetric tensor in dimension $n$ has $\frac {n(n-1)}2$ components.  You can choose the components above the main diagonal as you wish, the main diagonal has to be $0$ and the components below the main diagonal are the negatives of the ones above.  That gives $\frac {n(n-1)}2$ independent components.  If $n=3$, you can convert this to a(n axial) vector by taking the component of the vector to be the tensor element in the other two dimensions.  In $4$ dimensions an antisymmetric tensor has six independent components, so we can't view this as a vector which would have four.
