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An $r$-form is a totally antisymmetric $(0,r)$-tensor. Is there an analogous subset of $(r,0)$-tensors?

Additionally, is an $r$-form the same thing as an $r$-vector (e.g., is a 2-form the same as a bivector), or is an $r$-vector the dual of an $r$-form, in the same way that a vector is the dual of a covector. If so, does this define an analogous subset of $(r,0)$-tensors?

Alternatively, given a metric, can you define the dual of a $r$-form in the usual way that you would with a tensor (raising and lowering indices).

Edit:

I'm working in the context of manifolds, so a vector, $V_p$, at a point $p\in\mathcal{M}$ with chart $x:\mathcal{M}\to\mathbb{R}^m$ is:

$$ V_p \equiv [\gamma] \equiv \left.\frac{\mathrm{d}x^{\mu}\left[\gamma(t)\right]}{\mathrm{d}t}\frac{\partial}{\partial x^{\mu}}\right|_{p} $$

where $[\gamma]$ is an equivalence class of curves tangent at $p$, so $\forall\gamma_{1},\gamma_{2}\in[\gamma]$:

$$ \left.\frac{\mathrm{d}x^{\mu}\left[\gamma_{1}(t)\right]}{\mathrm{d}t}\right|_{p} = \left.\frac{\mathrm{d}x^{\mu}\left[\gamma_{2}(t)\right]}{\mathrm{d}t}\right|_{p} $$

The space of vectors at $p$ is the tangent space at $p$, $T_{p}\mathcal{M}$.

A covector, $A^p$, at a point $p$ with inverse chart $x^{-1}:\mathbb{R}^m \to \mathcal{M}$ is defined as:

$$ A^p \equiv [f] \equiv \left.\frac{\partial f\left[x^{-1\nu}(p)\right]}{\partial x^{\mu}}\right|_{p}\mathrm{d}x^{\mu} $$

where $[f]$ is the equivalence class of functions with the same gradient at $p$, so $\forall f_1,f_2\in[f]$:

$$ \left.\frac{\partial f_1\left[x^{-1\nu}(p)\right]}{\partial x^{\mu}}\right|_{p} = \left.\frac{\partial f_2\left[x^{-1\nu}(p)\right]}{\partial x^{\mu}}\right|_{p} $$

The space of covectors at $p$ is the cotangent space at $p$, $T_p^{*}\mathcal{M}$.

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  • $\begingroup$ Yes to your first question, no to your second - they are related by duality. What is your definition of a $(0,r) / (r,0)$ tensor and of an $r$-form / $r$ multivector? I'm asking because a detailed answer to your question depends on your initial definitions. $\endgroup$ – levap May 1 '17 at 16:43
  • $\begingroup$ Added definitions of vectors and covectors, tensors are just defined by the tensor product, and r-forms are antisymmetric tensors. I don't have a definition of a multivector, which is why I'm asking. $\endgroup$ – gautampk May 1 '17 at 17:05
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As your question is actually a question in linear algebra and not in manifold theory, it is best to uncouple the geometry. Let $V$ be a finite dimensional real vector space (in your case, $V = T_pM$). It seems that according to your definitions, a $(0,r)$-tensor on $V$ is an element of $(V^{*})^{\otimes r}$. By choosing a basis $\varphi^1,\dots,\varphi^n$ for $V^{*}$, any $(0,r)$ tensor $T$ can be written uniquely as

$$ T = T_{i_1,\dots,i_r} \varphi^{i_1} \otimes \dots \otimes \varphi^{i_r} $$

for real numbers $T_{i_1,\dots,i_r}$ which are called the components of $T$. An $r$-form then can be defined as a $(0,r)$ tensor of the form above whose components are totally antisymmetric (this can readily be seen to be independent of the basis chosen to represent $T$).

The same thing works for $(r,0)$ tensors and $r$ multivectors by replacing $V^{*}$ with $V$. An $(r,0)$ tensor $T$ is an element of $(V^{*})^{\otimes r}$ and if $e_1,\dots,e_n$ is a basis for $V$ then $T$ has the form

$$ T = T^{i_1 \dots i_r} e_1 \otimes \dots \otimes e_r. $$

An $r$ multivector can then be defined as a $(r,0)$ tensor whose components are totally antisymmetric. This is really not the best way to define $r$ multivectors or $r$ forms but since it seems that these are your definitions, we'll stick to that.

Finally, an $(0,r)$-tensor is not the same as an $(r,0)$ tensor just like a linear functional is not the same as a vector. However, just like the case of vectors and linear functionals, they are related by duality. An $(0,r)$ tensor can act on a $(r,0)$ tensor to yield a number and vice versa. This gives us a pairing

$$ (V^{*})^{\otimes r} \times (V)^{\otimes r} \rightarrow \mathbb{R} $$

which is in fact non-degenerate so we can identify $\left( (V^{*})^{\otimes r} \right)^{*}$ with $V^{\otimes r}$ (the dual of an $(0,r)$-tensor is an $(r,0)$-tensor).

Similarly, one can restrict/redefine the pairing and get a pairing between $r$ forms and $r$ multivectors so that $r$ forms and $r$ multivectors become dual to each other.

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  • $\begingroup$ Okay, great thanks. Out of curiosity, what would a better definition of r-forms be? This is the definition my lecturer starts with in his notes and then uses it to motivate the exterior product, but if there's a better equivalent definition then I'm sure it'll be useful to know. $\endgroup$ – gautampk May 1 '17 at 17:49

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