How are these two ways of thinking about the cross product related? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:15:04Z http://mathoverflow.net/feeds/question/33896 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33896/how-are-these-two-ways-of-thinking-about-the-cross-product-related How are these two ways of thinking about the cross product related? Qiaochu Yuan 2010-07-30T07:13:55Z 2010-07-30T11:31:57Z <p>I was always bothered by the definition of the cross product given in e.g. a calculus course because it's never made clear how one would go about defining the cross product in a coordinate-free manner. I now know, not one, but <em>two</em> ways of doing this, and I can't quite see how they're related:</p> <ul> <li>The cross product is the Lie bracket in the Lie algebra of $\text{SO}(3)$.</li> <li>The cross product is the Hodge star map $\Lambda^2(V) \to V$ where $V$ is an oriented $3$-dimensional real inner product space.</li> </ul> <p>Okay, so there's one obvious relation here: $V$ has automorphism group $\text{SO}(3)$. But for some reason I can't figure out where to go from here. A good starting point would be to exhibit a canonical isomorphism between an oriented $3$-dimensional inner product space $V$ and the Lie algebra of $\text{Aut}(V)$. Maybe this is obvious. In any case, I would appreciate some clarification.</p> http://mathoverflow.net/questions/33896/how-are-these-two-ways-of-thinking-about-the-cross-product-related/33903#33903 Answer by Joel Fine for How are these two ways of thinking about the cross product related? Joel Fine 2010-07-30T09:03:50Z 2010-07-30T10:27:45Z <p>To expand on Victork Protsak's comment, if V is an n-dimensional real vector space with inner-product, the inner-product gives an isomorphism $V\to V^*$ and hence $V\otimes V \to \mathrm{End}(V)$. Under this isomorphism, $\Lambda^2(V)$ is identified with skew-adjoint endomorphisms of $V$, which is precisely the Lie algebra $\mathfrak{so}(V)$.</p> <p>In the case dim V =3, the Hodge star gives an isomorphism $\Lambda^2(V) \to V$ and so in total we see that $V$ is canonically isomorphic to $\mathfrak{so}(V)$. A more direct way to see this isomorphism is to send the vector $v \in V$ to the generator of the right-handed rotation about the axis in the direction of $v$ with speed $|v|$. </p> <p>The use of the phrase "right-handed" makes it clear that in order to identify $V$ and $\mathfrak{so}(V)$ we have used an orientation on $V$; indeed, you need that for the Hodge star. What is interesting is that if you reverse the orientation on $V$, the map to $\mathfrak{so}(V)$ changes sign. This means that what ever orientation you chose on $V$, the push-forward to $\mathfrak{so}(V)$ is the same. Conclusion: $\mathfrak{so}(3)$ is <i>naturally oriented</i>. This is analogous to the natural orientation on $\mathbb{C}$. A more prosaic way to describe the orientation is to pick two independent elements $x,y \in \mathfrak{so}(3)$ and then use $[x,y]$ to complete them to an oriented basis. (Of course, you then need to check that this doesn't depend on your choice of $x,y$.)</p> http://mathoverflow.net/questions/33896/how-are-these-two-ways-of-thinking-about-the-cross-product-related/33907#33907 Answer by Christian Blatter for How are these two ways of thinking about the cross product related? Christian Blatter 2010-07-30T11:31:57Z 2010-07-30T11:31:57Z <p>Let $\varepsilon( )$ be the volume form in $\mathbb R^3$. For given vectors ${\bf p}$ and ${\bf q}$ the function $f:{\bf x}\mapsto\varepsilon({\bf p},{\bf q},{\bf x})$ is a linear functional and so is represented by a vector ${\bf r}\in\mathbb R^3$, i.e., one has $f({\bf x})=\langle{\bf r},{\bf x}\rangle$. This vector ${\bf r}$ depends in a skew bilinear way from ${\bf p}$ and ${\bf q}$ and is called the $vector\ product$ of ${\bf p}$ and ${\bf q}$.</p>