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Let $A$ = {triangles in $\mathbb{R^2}$}. We can let $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$ be the vertices of the triangle. The group $GL(2,\mathbb{R})$ acts on $A$ by acting on the vectors of the vertices which form the triangle. $GL(2,\mathbb{R})$ acts on $\mathbb{R}$ via the action: $y \in \mathbb{R}$, $B \in GL(2,\mathbb{R})$ then $B(y) = |\det(B)|y$. Now we let $f: A \rightarrow \mathbb{R}$ which maps a triangle in $A$ to its area. We want to show that $f$ is equivariant with respect to these group actions.

Solution attempt: Let $t \in X$ and $A \in GL(2, \mathbb{R})$. $A(f(t))=|\det(A)|f(t) $ where $f(t)$ is the area of triangle $t$. We can let the area of $t$ be represented by a 3x3 determinant from linear algebra. From here I thought of use determinant properties, but we cannot multiply a 2x2 to a 3x3 matrix.

user77404
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1 Answers1

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Hint: we may compute the area of a triangle via a $2\times2$ determinant of two displacement vectors.

anon
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  • I think I know what your trying to say. We can split the 3x3 determinant of the area into a sum of three separate 2x2 determinants and then go from there? – user77404 Aug 06 '13 at 00:51
  • No. Form two displacement vectors to represent two sides of the triangle. Put them together to form a $2\times2$ matrix. Halve the absolute value of the determinant and you have the area of the triangle. – anon Aug 06 '13 at 00:52
  • I think I remember this from linear algebra. When we have a parallelogram we can find the area of it by use the two displacement vectors. For triangle maybe we could divide by 2, it is along those lines, could I have a link to maybe a general formula? – user77404 Aug 06 '13 at 00:56
  • @user77404 Yes (note that "halve" means "divide by 2"). You've just said the general formula yourself in your comment, but here's a link to some mathematics behind it. Although by 'general formula' perhaps you're referring to the fact that the determinant of a $n\times n$ matrix is the hypervolume of the parallelotope formed by the $n$ column vectors? In this case too, the general formula is right there in the previous sentence. – anon Aug 06 '13 at 00:59
  • Great, and from that we could simplify what we have above into 1/2(det(AT)) where T is the matrix that determined area of T and from this we can conclude this also equals f(A(t)) for equivariant. But the only thing is we are not told what action $Gl(2,R)$ has on the vertices, so will we be sure that f(A(t)) = 1/2(det(AT))? – user77404 Aug 06 '13 at 01:08
  • It is assumed that GL(2,R) acts on the vertices by applying linear transformations to the vectors that represent the vertices. (This is the action general linear groups always come equipped with by default.) – anon Aug 06 '13 at 01:09