Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:56:53Z http://mathoverflow.net/feeds/question/66444 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66444/generalization-of-scalar-product-to-k-dimensional-subspaces-as-opposed-to-1-dime Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors) Szabolcs 2011-05-30T13:02:47Z 2011-05-30T17:21:03Z <p>This came up in a practical problem (physics).</p> <p>In the following, we work with real numbers only, and consider every vector to be normalized to 1.</p> <p>To find how "similar" two vectors are (actually, two lines passing through the origin, I don't care about the direction), one can use the scalar product. If the two lines are the same, then the scalar product of the corresponding vectors is 1 or -1. If they are "similar", it's close to 1. If they're perpendicular, it's 0.</p> <p>I need a generalization of this "similarity measure" to $k$-dimensional subspaces.</p> <p>For example, for $k=2$ I have the following problem: I have vectors $a_1 \perp a_2$ which define a plane in an $n$-dimensional space, and vectors $b_1 \perp b_2$. I need a measure that 1. tells me how close these planes are to each other, in terms of $a_1, a_2, b_1, b_2$. is a generalization of the simple scalar product (i.e. it's the same as the scalar product for $k=1$).</p> http://mathoverflow.net/questions/66444/generalization-of-scalar-product-to-k-dimensional-subspaces-as-opposed-to-1-dime/66446#66446 Answer by Qiaochu Yuan for Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors) Qiaochu Yuan 2011-05-30T13:30:38Z 2011-05-30T13:30:38Z <p>Let $V$ be an inner product space. The tensor product $V^{\otimes k}$ is equipped with an inner product induced from that of $V$ (obtained by contracting via the inner product $k$ times). Concretely, if $e_1, ... e_n$ is an orthonormal basis for $V$ then all sequences of tensor products of the $e_i$ form an orthonormal basis for $V^{\otimes k}$. Now, $V^{\otimes k}$ has a canonical subspace, the space of <a href="http://en.wikipedia.org/wiki/Antisymmetric_tensor" rel="nofollow">antisymmetric tensors</a> $\text{Alt}^k(V)$ (not to be confused with the exterior power, which is a quotient of $V^{\otimes k}$ rather than a subspace), and the above inner product induces an inner product on antisymmetric tensors. Finally, given an orthonormal basis $f_1, ... f_k$ of a $k$-dimensional subspace $W$ of $V$, the antisymmetrization</p> <p>$$f_1 \wedge ... \wedge f_k = \frac{1}{\sqrt{k!}} \sum_{\sigma \in S_k} \text{sgn}(\sigma) f_{\sigma(1)} \otimes ... \otimes f_{\sigma(k)}$$</p> <p>associates to $W$ an element of $\text{Alt}^k(V)$ invariant up to oriented change of orthonormal basis. I <em>think</em> the inner product of these elements has the property you want. </p> http://mathoverflow.net/questions/66444/generalization-of-scalar-product-to-k-dimensional-subspaces-as-opposed-to-1-dime/66447#66447 Answer by Spiro Karigiannis for Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors) Spiro Karigiannis 2011-05-30T13:31:15Z 2011-05-30T13:31:15Z <p>Let $V$ be an $n$-dimensional real vector space with an inner product $\langle \cdot,\cdot \rangle$. Then $\Lambda^k (V)$, the $k^{\text{th}}$ exterior product of $V$, inherits an induced inner product defined as follows.</p> <p>If $\alpha = v_1 \wedge \cdots \wedge v_k$ and $\beta = w_1 \wedge \cdots \wedge w_k$, then $\langle \alpha , \beta \rangle = \det \langle v_i , w_j \rangle$ on decomposable vectors, and then extend by linearity to all of $\Lambda^k(V)$. Now a $k$-dimensional subspace $P$ of $V$ can be represented by a decomposable element $\alpha_P = v_1 \wedge \cdots \wedge v_k$ of $\Lambda^k(V)$, where $v_1, \ldots, v_k$ is a basis for $P$. If the basis is taken to be oriented and orthonormal, then the element $\alpha_P$ is uniquely determined. (Otherwise it is only determined up to a non-zero real scalar, so in general it defines a point in the projectivization $\mathbb P (\Lambda^k(V))$. Since this is a real inner product on a finite-dimensional vector space, it does indeed give you a measure of "how far" two vectors are from being parallel, or in this case, how far two $k$-dimensional subspaces are from being the same.</p> <p>If $k = 2$ and $a_1, a_2$ and $b_1, b_2$ are two oriented orthonormal bases for two planes $P_a$ and $P_b$ in $V$, then</p> <p>$$\langle P_a, P_b \rangle = \det{\langle a_i, b_j \rangle} = \langle a_1, b_1 \rangle \langle a_2, b_2 \rangle - \langle a_1, b_2 \rangle \langle a_2, b_1 \rangle.$$</p> <p>You can search for "induced inner product on exterior algebra" or look at any book on differential forms for more details.</p> http://mathoverflow.net/questions/66444/generalization-of-scalar-product-to-k-dimensional-subspaces-as-opposed-to-1-dime/66448#66448 Answer by robot for Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors) robot 2011-05-30T13:37:10Z 2011-05-30T17:21:03Z <p>The set of all $k$-dimensional subspaces of $\mathbb{R}^n$ is called <a href="http://en.wikipedia.org/wiki/Grassmannian" rel="nofollow">Grassmannian</a> $\mathrm{Gr}(k,n)$ and can be given a structure of a smooth manifold. That gives you a topological notion of nearness, but there is more. There is a homogeneous Riemannian metric on this manifold that can be expressed quite explicitly - details can be found in this <a href="http://www.springerlink.com/content/vr1578pp26q3001t/" rel="nofollow">article</a> by Yuri Neretin. Or you can try this <a href="http://www.springerlink.com/content/hq88q716l525511u/" rel="nofollow">one</a> by Sheng Jiang which may be more accessible for you.</p> http://mathoverflow.net/questions/66444/generalization-of-scalar-product-to-k-dimensional-subspaces-as-opposed-to-1-dime/66450#66450 Answer by Steve Flammia for Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors) Steve Flammia 2011-05-30T13:55:28Z 2011-05-30T13:55:28Z <p>There is a very simple way to do this which is to consider the projector onto each subspace, $\Pi_A$ and $\Pi_B$, where we have $$\Pi_A = \sum_j a_j a_j^T$$ and similarly for $\Pi_B$. Since all of your vectors spanning your subspace are orthogonal and normalized, this is indeed a rank-$k$ projector. Then you can just consider the Hilbert-Schmidt inner product, $$\langle A , B \rangle = \mathrm{Tr}(A^T B)$$ and this will give you a measure of how similar the two spaces are. If you first divide each projector by the square root of it's rank, then this quantity is normalized so that it is 1 if and only if the two subspaces are equal. </p>