The Unapologetic Mathematician

Mathematics for the interested outsider

The Determinant

Let’s look at the dimensions of antisymmetric tensor spaces. We worked out that if V has dimension d, then the space of antisymmetric tensors with n tensorands has dimension


One thing should leap out about this: if n is greater than d, then the dimension formula breaks down. This is connected with the fact that at that point we can’t pick any n-tuples without repetition from d basis vectors.

So what happens right before everything breaks down? If n=d, 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 V^{\otimes d} is both a representation of \mathrm{GL}(V) and a representation of S_d, 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 \mathrm{GL}(V) action. Thus we have a one-dimensional representation of \mathrm{GL}(V), 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 V 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 \mathrm{GL}(V) is defined completely without reference to any basis of V. 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.

December 31, 2008 - Posted by | Algebra, Linear Algebra, Representation Theory


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