Is there a name for this map induced by bilinear forms? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:52:49Z http://mathoverflow.net/feeds/question/73885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73885/is-there-a-name-for-this-map-induced-by-bilinear-forms Is there a name for this map induced by bilinear forms? Colin Tan 2011-08-28T07:46:38Z 2011-08-28T20:27:57Z <p>Let $V$ be a real vector space. A bilinear form $\langle \rangle:V\times V\to {\mathbb{R}}$ induces a linear functional $\theta$ on the tensor product $V\otimes V$ given by sending the finite sum $\sum_i v_i\otimes w_i$ to $\sum_i \langle v_i,w_i\rangle$.</p> <p>Is there a name for this induced linear functional?</p> <p>In addition, if the bilinear form is symmetric, then this linear functional $\theta$ respects the natural involution on $V\otimes V$. That is $\theta(v\otimes w)=\theta(w\otimes v)$.</p> http://mathoverflow.net/questions/73885/is-there-a-name-for-this-map-induced-by-bilinear-forms/73887#73887 Answer by kostja for Is there a name for this map induced by bilinear forms? kostja 2011-08-28T08:03:22Z 2011-08-28T09:07:09Z <p>Hello,</p> <p>I don't know a specific name for that, but I would call it associated. I wouldn't call it induced, because the map $Bil(V) \to (V \otimes V)^*$ is one-to-one, since every linear $T \in (V\otimes V)^*$ induces a bilinear form on $V$ by sending $(v,w) \mapsto T(v\otimes w)$ and this clearly is the inverse.</p> <p>Kind regards Konstantin</p> http://mathoverflow.net/questions/73885/is-there-a-name-for-this-map-induced-by-bilinear-forms/73904#73904 Answer by Ray Ryan for Is there a name for this map induced by bilinear forms? Ray Ryan 2011-08-28T17:32:03Z 2011-08-28T17:32:03Z <p>And symmetry of bilinear forms can be encoded in the tensor product: if we define the symmetric tensor product $V\otimes_s V$ to be the subspace of $V \otimes V$ spanned by the "symmetric tensors" $v\otimes_s w = (v\otimes w + w\otimes v)/2$, then there is a canonical bijection between linear forms on $V\otimes_s V$ and symmetric bilinear forms on $V\times V$.</p> http://mathoverflow.net/questions/73885/is-there-a-name-for-this-map-induced-by-bilinear-forms/73907#73907 Answer by paul garrett for Is there a name for this map induced by bilinear forms? paul garrett 2011-08-28T17:52:24Z 2011-08-28T17:52:24Z <p>There may be some interest in a variant: the natural map $V\otimes V^{\star}$ to $End(V)$ hits all finite-rank endomorphisms of $V$, and the bilinear map $V\times V^\star\to k$ "induces" <em>trace</em> on finite-rank endos. Thus, some name like "trace" is quite nearby to the literal question.</p> http://mathoverflow.net/questions/73885/is-there-a-name-for-this-map-induced-by-bilinear-forms/73908#73908 Answer by Ryan Reich for Is there a name for this map induced by bilinear forms? Ryan Reich 2011-08-28T17:55:27Z 2011-08-28T17:55:27Z <p>The correspondence you describe is part of the definition of the tensor product: $V \otimes W$ is defined to have the universal property that for any $U$, we have $$\operatorname{Hom}(V \otimes W, U) = \operatorname{Bil}(V \times W, U).$$ I wouldn't even give it a different name: the bilinear form is the same as the map out of the tensor product. </p> http://mathoverflow.net/questions/73885/is-there-a-name-for-this-map-induced-by-bilinear-forms/73913#73913 Answer by Bill Cook for Is there a name for this map induced by bilinear forms? Bill Cook 2011-08-28T20:27:57Z 2011-08-28T20:27:57Z <p>A bilinear form on $V$ (if non-degenerate) lets you identify $V$ with $V^\star$: $v \mapsto \langle v, \cdot \rangle$</p> <p>In this case, your map is just a <b>contraction</b> of the identity map on $V^\star \otimes V$ (which, considering our identification, is the same as the identity on $V \otimes V$).</p> <p><a href="http://en.wikipedia.org/wiki/Tensor_contraction" rel="nofollow">http://en.wikipedia.org/wiki/Tensor_contraction</a></p>