Tagged Questions

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0answers
50 views

Operations on vector spaces

Is there a symmetrical analogue of the grassmann tensor products? Is it the polyadic tensor product? Are such symmetrical products used anywhere in differential geometry?
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0answers
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Differentiable manifolds, Serge Lang

I have started reading "Introduction to differentiable manifolds" by Serge Lang. In this book, Lang takes a different approach, by immediately introducing manifolds on arbitrary Banach spaces. His ...
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1answer
43 views

Help with algebraic manipulation to prove that $\Omega (M)$ constructed from $\Omega^1 (M)$ forms an algebra over $C^\infty (M)$

In Baez´s Gauge Theories, Knots and Gravity he states that the differential forms on a n dimensional manifold M, $\Omega (M) = {\bigoplus}_p \Omega^p (M)$, constructed from $\Omega^1(M)$ and the ...
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1answer
176 views

Geometric meaning of Gram determinant

Let $v_1,v_2$ be vectors in $\mathbb{R}^4$. Let $M$ be the $2\times 4$ matrix with rows $v_1,v_2$ in this order. The Gram determinant of $M$ is defined as the determinant of the $2\times 2$ matrix ...
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1answer
192 views

how to understand the tensor product canonical line bundle $\otimes$ dual bundle

Suppose we have a Riemann surface $M$ together with a holomorphic vector bundle $E \to M$ of rank n. let $K$ denote the canonical line bundle and let $E^*$ denote the dual bundle I am trying to ...
3
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1answer
87 views

Linear independence regarding Exterior Power .

I have been trying to learn the proof of dimension of exterior power from this text : http://www.thehcmr.org/issue1_2/poincare_lemma.pdf.( Page 16) I am not able to understand the part of linear ...
5
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1answer
337 views

Covectors $\omega^1, …, \omega^k$ are linearly dependent iff their wedge product is zero

How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?
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2answers
184 views

Wedge product and linear subspace

I am trying to understand the relationship between the wedge product and linear subspace. Let $e_1,\cdots, e_4$ be the standard basis of $\mathbb{R}^4$. The wedge product $$(e_1+2e_2)\wedge ...
3
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1answer
260 views

Universal Definition for Pullback

The concept of "pullback" has several definitions depending on the context in which it is applied, e.g., smooth functions on manifolds, differential forms, multilinear forms and so forth. See, for ...
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2answers
561 views

Are “differential forms” an algebraic approach to multivariable calculus?

I am recently learning some basic differential geometry. As I understand, differential forms provide a neat way to deal with the topics in calculus such as Stoke's theorem. In order to define the ...
0
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1answer
170 views

Canonical Isomorphism Between $\Omega^2(\mathbb{R^3})$ and $\mathbb{R^3}$?

Let $\Omega^2(\mathbb{R}^3)$ represent the collection of differential 2-forms on $\mathbb{R}^3$. For this space we take as an (ordered) basis $\{dx \wedge dy, dx \wedge dz, dy \wedge dz\}$. First ...
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4answers
705 views

Concrete Example Illustrating the Interior Product

Let $V$ be a finite-dimensional vector space, let $v \in V$ and let $\omega$ be an alternating $k$-tensor on $V$, i.e., $\omega \in \Lambda^{k}(V)$. Then, the interior product of $v$ with $w$, denoted ...
5
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1answer
154 views

Extension of Riemannian Metric to Higher Forms

I've been reading about Riemannian manifolds, and have come across a comment that says that for a metric $g$ on an $N$-dimensional manifold $M$, considered as a bilinear map $$ g:\Omega^1(M) \times ...