Let’s look at the dimensions of antisymmetric tensor spaces. We worked out that if has dimension , then the space of antisymmetric tensors with tensorands has dimension
One thing should leap out about this: if is greater than , then the dimension formula breaks down. This is connected with the fact that at that point we can’t pick any -tuples without repetition from basis vectors.
So what happens right before everything breaks down? If , then we find
There’s only one independent antisymmetric tensor of this type, and so we have a one-dimensional vector space. But remember that this isn’t just a vector space. The tensor power is both a representation of and a representation of , which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the action. Thus we have a one-dimensional representation of , which we call the determinant representation.
I want to pause here and point out something that’s extremely important. We’ve mentioned a basis for in the process of calculating the dimension of this space, but the space itself was defined without reference to such a basis. Similarly, the representation of any element of is defined completely without reference to any basis of . It needs only the abstract vector space itself to be defined. Calculating the determinant of a linear transformation, though, is a different story. We’ll use a basis to calculate it, but as we’ve just said the particular choice of a basis won’t matter in the slightest to the answer we get. We’d get the same answer no matter what basis we chose.