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Let $V$ be a real vector space of dimension $2$, and let $\langle\ \ ,\ \ \rangle$ be an inner product on $V$. Define $f:V^4 \to \mathbb{R}$ by $$f(x,y,z,w):=\langle x,y \rangle \langle z,w \rangle- \langle x,w \rangle \langle z,y \rangle$$

Show there's a skew-symmetric bilinear form $g:V^2 \to \mathbb{R}$ such that $$f(x,y,z,w)=g(x,z)g(y,w)$$

My thought: Let $z=y,w=x$ to get $f(x,y,y,x)= \langle x,y \rangle ^2-\langle x,x \rangle \langle y,y \rangle$. If $g$ exists, then $f(x,y,y,x)=g(x,y)g(y,x)=-g(x,y)^2$ (notice that $g$ is skew-symetric). Equating both identities we get $g(x,y)^2= -\langle x,y \rangle ^2+\langle x,x \rangle \langle y,y \rangle$. Here is where I can't proceed any more...

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    Hint: How many linear independent skew-symmetric bilinear forms are there on a vector space of dimension $2$?2012-08-23
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    Choose an orthonormal basis. What is $f(e_1,e_1,e_2,e_2)$?2012-08-23
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    It's 1, so $g(x,y)=\det (x,y)$.. Thanks! Just another query, is $\det(x,y)$ the standard notation in abstract algebra for a basis-independent determinant?2012-08-23
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    actually, the standard way to write this is $e^1\wedge e^2$. $e^i$ are the covectors (linear forms) dual to the basis $e_1$, $e_2$ (that is, $e^i(e_j)=\delta^i_j$ where the right hand side is the [Kronecker delta](https://en.wikipedia.org/wiki/Kronecker_delta), and $\wedge$ is the exterior product or wedge product, which is used to build $n$-forms (skew-symmetric $n$-linear functions on the vector space). Especially, if $\alpha$ and $\beta$ are $1$-forms (linear functions), $(\alpha\wedge\beta)(u,v)=\alpha(u)\beta(v)-\beta(u)\alpha(v)$.2012-08-24
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    Of course $e^1\wedge e^2$ is the "determinant form" (the proper name is "volume form) only for the two-dimensional vector space (the definition is of course general, but for more dimensions, this doesn't give the determinant). For a general $n$-dimensional vector space, it would be $e^1\wedge e^2\wedge\dots\wedge e^n$.2012-08-24
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    Thanks! I understand that picking an orthonomal basis is necessary to construct $g$ explicity, but I wonder at the moment is there a "basis independent" way to deduce the existence of $g$? (Is it possible to do this just by using your hint at the start that the dimension of $\Lambda^{3}V^{*}$ is 1?) If there is then one can avoid the "determinant form" and it may be easily generalised to higher dimensions.2012-08-24
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    One could start with noting that $f$ is skew-symmetric in the first and third argument, and also in the second and fourth argument. I think that this (plus linearity of $f$ in all arguments) should be enough to conclude that $f$ is the product of two $2$-forms. Be $g_1$ and $g_2$ $2$-forms whose product is $f$. Due to dimension 1, $g_2=\lambda g_1$ for some number $\lambda\ne 0$. Therefore if you define $g=\sqrt{\lvert\lambda\rvert}g_1$, you have $f(x,y,z,w) = \pm g(x,z)g(y,w)$. To get the sign, you only have to check that $f(x,x,z,z)$ is positive.2012-08-24
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    True, and this can be naturally generalised to any rank $2n$ tensor product which has similar form to $f$. Thanks for all the comments!2012-08-24
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    @celtschk Please consider converting your comments into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-22
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    @JulianKuelshammer: OK, I've now expanded the comments into a full answer (which actually got quite long because I expanded all the hints into the actual calculations).2013-06-22

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