# Bilinear Map on Exterior Algebra That Gives Determinant

How do I see that for a $$K$$-vector space $$V$$ the map

$$\bigwedge^d(V^*) \times \bigwedge^d(V) \rightarrow K, (f_1 \wedge ... \wedge f_d, x_1 \wedge ... \wedge x_d) \mapsto det(f_i(x_i)_{i,j})$$

is bilinear?

• The problem is to show / know it is welll defined. If we know that the result is obvious, we have linearity in first variable $f=(f_1,f_2,\dots,f_d)$ (well, the map factors through the wedge product if it is well defined) since the determinant is multillinear w.r..t to its rows/columns,. The same with $x= (x_1,x_2,\dots, x_d)$ instead of $f$. – dan_fulea Jun 26 at 22:17
• @dan_fulea Then how do I show well-definedness? – Chaser01 Jun 26 at 22:26

I try to write an answer that should clear the definition of the map in the OP. By definition, it is a (multi)linear map. The main instrument is using "universality" when working with elements in the category of vector spaces and (multi)linear applications. (This would not fit as a comment, and it would be hard to type without markup control.)

(1) Let us fix some field $$K$$ (of characteristic $$\ne 2$$, or maybe even $$=0$$ to exclude any problems with the definition of the wedge space).

We work in the category of vector spaces over $$V$$.

Functorially, if $$f,g$$ are (linear) maps $$V\to V'$$ and $$W\to W'$$, then $$f\otimes g$$ is a (bi)linear map $$V\times W\to V'\times W'$$. Same is valid for more tensor factors.

(2) Consider now "the $$V$$" from the OP. We have an evaluation map $$V^*\otimes V\to K$$.

Using it we can define for a fixed pair $$(i_0,j_0)$$ the map $$\left(\times_{i=1}^dV^*\right)\ \otimes\ \left(\times_{j=1}^dV\right) \to K\ ,$$ $$(f_1,f_2,\dots,f_d)\otimes(x_1,x_2,\dots,x_d) \to f_{i_0}(x_{j_0})\ .$$

(3) Putting together all the above maps for all possible values of $$(i_0, j_0)$$, so that the image space is a space of matrices $$d\times d$$, we have a map By repeating it in $$d$$ tensor parts, we also have a map: $$\left(\times_{i=1}^dV^*\right)\ \otimes\ \left(\times_{j=1}^dV\right) \to M_{d\times d}(K)\ ,$$ $$(f_1,f_2,\dots,f_d)\otimes(x_1,x_2,\dots,x_d) \to \Big[\ f_{i_0}(x_{j_0})\ \Big]_{1\le i_0,j_0\le d}\ .$$ As it is so far, this map is linear (in each component), but it is not "balanced", i.e. we cannot move a scalar from one $$f$$-component to an other one, or from an $$x$$-component to an other one.

(4) Consider now the composition $$\left(\times_{i=1}^dV^*\right)\ \otimes\ \left(\times_{j=1}^dV\right) \to M_{d\times d}(K) \overset\det\longrightarrow K\ .$$ This composition is now balanced by the properties of the determinant. For instance, $$(af_1,f_2,\dots,f_d)\otimes x$$ and/or $$f\otimes(ax_1,x_2,\dots,x_d)$$ is mapped via (3) to the matrix obtained from the one for $$f\otimes x=(f_1,f_2,\dots,f_d)\otimes (x_1,x_2,\dots,x_d)$$ by multiplying the first matrix row/column with the scalar $$a\in K$$.

Applying the $$\det$$, $$a$$ becomes now a factor of the result.

The same computation can be done when $$a$$ is on an other component position of $$f$$ and/or of $$x$$.

So the balancing properties is valid after applying $$\det$$.

(5) The balancing implies we have an induced map $$\bar\Phi$$: $$\bar\Phi\ :\ \left(\bigotimes_{i=1}^dV^*\right)\ \otimes\ \left(\bigotimes_{j=1}^dV\right) \to K\ .$$ On elements $$f_1\otimes f_2\otimes\dots\otimes f_d$$ algebraically tensored with $$x_1\otimes x_2\otimes\dots\otimes x_d$$ it is defined by lifting this to (linear combinations of) $$(f_1,f_2,\dots,f_3)\otimes(x_1,x_2,\dots,x_d)$$ and applying $$\Phi$$ (followed by linear assembly).

The result does not depend on the lifts. Every relation that has to be tested for $$\bar\phi$$ has an equivalent pendant at the level of $$\Phi$$.

(6) It remains to observe that $$\bar\Phi$$ is alternating in its $$f$$-components, and also in its $$x$$-components. We only need to show this at the level of $$\Phi$$.

So we have to compare first the result of applying $$\Phi$$ on the elements

$$(\color{blue}{f_1,f_2},\dots,f_d)\otimes(x_1,x_2,\dots,x_d)$$ and respectively

$$(\color{blue}{f_2,f_1},\dots,f_d)\otimes(x_1,x_2,\dots,x_d)$$

(and on all other cases of a change implemented by a transposition of two indices).

We apply $$\Phi$$ on the above two elements, the intermediate matrix station delivers two matrices with interchanged first and second rows, further applying $$\det$$ leads to a sign difference. In this case and in the other cases of using a transposition of indices of the $$f$$-component

This shows the alternating relation for the $$f$$-components.

The similar argument applied for the comparison of $$\Phi$$-values on

$$(f_1,f_2,\dots,f_d)\otimes(\color{blue}{x_1,x_2},\dots,x_d)$$ and respectively

$$(f_1,f_2,\dots,f_d)\otimes(\color{blue}{x_2,x_1},\dots,x_d)$$

and on the values in the more general case, when we are using a transposition $$(j_1,j_2)$$ instead of $$(1,2)$$ as above,

is leading to the comparison of two determinants with matrices two exchanged columns, and again we deduce the alternation relation, this time on the $$x$$-components.

(7) We can thus factorize through the wedge-product, getting a final map $$\hat\Phi$$: $$\hat\Phi\ :\ \left(\wedge_{i=1}^dV^*\right)\ \otimes\ \left(\wedge_{j=1}^dV\right) \to K\ .$$ (The wedge product can be realized in characteristic zero either als subobject, or as a quotient of the tensor product. The factorization makes sense when the quotient is taken, and the map is already alternating.)

• Thank you very much for this detailed answer, it really helps a lot! But this map is defined on $\bigwedge^d(V^*) \otimes \bigwedge^d(V)$ instead of $\bigwedge^d(V^*) \times \bigwedge^d(V)$, isn't it? – Chaser01 Jun 27 at 15:20

The given formula is certainly giving a well defined mapping $$\varphi:V^*\times\dots\times V^*\ \times\ V\times\dots\times V \longrightarrow K$$ Fixing all but one arguments makes the matrix of the determinant varying linearly in one row or column.
This shows that $$\varphi$$ is multilinear, so that it factors through $$V^*\otimes\dots\otimes V^*\ \otimes\ V\otimes\dots\otimes V \longrightarrow K$$ Finally, if $$f_i=f_j$$ [or $$x_i=x_j$$] with $$i\ne j$$, the matrix of the determinant will have two identical rows [columns], which shows that restricting to the first [second] $$d$$ variables gives an alternating multilinear map, and hence it factors through $$(V^*\land\dots\land V^*)\ \otimes\ (V\land\dots\land V)\ .$$

• Thank you very much, that helps me a lot! Just one question: I was searching for a map from the Cartesian product of $\bigwedge)V^*)$ and $\bigwedge)V)$ instead of one from their tensor product, how can we fix that? – Chaser01 Jun 27 at 11:37
• We have the following sequence: $(\times^d V^*)\times(\times^d V) \ \to\ (\otimes^d V^*) \times(\otimes^d V) \ \to\ (\otimes^d V^*) \otimes(\otimes^d V) \ \to\ (\land^d V^*) \otimes(\land^d V) \ \to\ K$. – Berci Jun 27 at 18:21
• Ahh you rather want this line: $(\otimes^d V^*) \times(\otimes^d V) \ \to\ (\land^d V^*) \times(\land^d V) \ \to\ (\land^d V^*) \otimes(\land^d V)\ \to \ K$. – Berci Jun 27 at 18:26