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Set of Covariant Tensors ... Browder Sections 12.7 and 12.8 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 12: Multilinear Algebra ... ...

I need help in order to fully understand the definition and nature of the set of covariant tensors of rank \(\displaystyle r\) ... as described in Browder, Section 12.7 and 12.8 ...


Te relevant text reads as follows:



Browder - 1 -  Sections 12.7 and 12.8  ... ... PART 1 .png
Browder - 2 -  Sections 12.7 and 12.8  ... ... PART 2 .png



My questions related to the above text are as follows:



Question 1

Given that \(\displaystyle T^r\) is the set of all (multilinear) maps from \(\displaystyle V^r\) to \(\displaystyle \mathbb{R}\) can we conclude that \(\displaystyle T^r\) is equal to the dual space of \(\displaystyle V^r\), visually \(\displaystyle (V^r)^*\) ... ?



Question 2

In Section 12.8 Browder writes the following ...

\(\displaystyle T^r = V^* \otimes \ ... \ \otimes V^*\) (r times)


What meaning can we give to this notation ... what clues does it give us about the nature of \(\displaystyle T^r\) ... indeed why is Browder mentioning/using this notation ... ?



Hope someone can help ...

Peter
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
255
Hi Peter ,

These are deep, insightful considerations that truly highlight your attention to detail. I think these are great questions because it is easy to overlook the subtlety regarding these topics.

Question 1

Given that \(\displaystyle T^r\) is the set of all (multilinear) maps from \(\displaystyle V^r\) to \(\displaystyle \mathbb{R}\) can we conclude that \(\displaystyle T^r\) is equal to the dual space of \(\displaystyle V^r\), visually \(\displaystyle (V^r)^*\) ... ?
Short Answer: No, $T^{r}\neq (V^{r})^{*}.$

Long Answer: The intuition to try and link $T^{r}$ with $(V^{r})^{*}$ is reasonable and tempting. Unfortunately, this is not the case.


Counterexample 1: Linear Functions Need Not be Multilinear

Consider $V=\mathbb{R}$, and let $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ be defined by $f(x,y)= 2x +3y.$ This is an example of a linear function on $\mathbb{R}^{2}=\mathbb{R}\times\mathbb{R}$ that is not multilinear.


Counterexample 2a: Multilinear Functions Need Not be Linear

Consider $V=\mathbb{R}^{2},$ and let $f:\mathbb{R}^{2}\times \mathbb{R}^{2}\rightarrow\mathbb{R}$ be defined to be the Euclidean inner product; i.e., $f({\bf x}, {\bf y}) = \langle {\bf x}, {\bf y}\rangle.$ This is an example of a multilinear function that is not linear because, using multilinearity, $$f({\bf x}_{1}+{\bf x}_{2}, {\bf y}_{1}+{\bf y}_{2})=f({\bf x}_{1},{\bf y}_{1})+f({\bf x}_{1},{\bf y}_{2})+f({\bf x}_{2},{\bf y}_{1})+f({\bf x}_{2},{\bf y}_{2}).$$ However, linearity would require $$f({\bf x}_{1}+{\bf x}_{2}, {\bf y}_{1}+{\bf y}_{2})= f({\bf x}_{1}, {\bf y}_{1}) + f({\bf x}_{2},{\bf y}_{2}).$$ The presence of the mixed terms $f({\bf x}_{1}, {\bf y}_{2})$ and $f({\bf x}_{2}, {\bf y}_{1})$ is the essence of the difference between multilinear and linear functions on products of vector spaces.


Counterexample 2b: Multilinear Functions Need Not be Linear

Let $V=\mathbb{R}^{n}$, and define $f:\underbrace{\mathbb{R}^{n}\times\cdots\times\mathbb{R}^{n}}_{n-\text{copies}}$ to be the determinant function; i.e., $f({\bf v}_{1},\ldots, {\bf v}_{n})=\text{Det}[{\bf v}_{1},\ldots, {\bf v}_{n}].$ It is know from the properties of determinants that $f$ is multilinear (see https://en.wikipedia.org/wiki/Determinant), but that $f$ is not linear because, in general, $\text{Det}(A+B)\neq \text{Det}(A)+\text{Det}(B)$.


Standard Multilinear Function Examples

The archetype of multilinear functions are the Euclidean inner product on $\mathbb{R}^{n}\times\mathbb{R}^{n}$, the cross product on $\mathbb{R}^{3}\times\mathbb{R}^{3}$, and the determinant on $\underbrace{\mathbb{R}^{n}\times\cdots\times\mathbb{R}^{n}}_{n-\text{copies}}.$


Question 2

In Section 12.8 Browder writes the following ...

\(\displaystyle T^r = V^* \otimes \ ... \ \otimes V^*\) (r times)


What meaning can we give to this notation ... what clues does it give us about the nature of \(\displaystyle T^r\) ... indeed why is Browder mentioning/using this notation ... ?
Short Answer: $(V^{r})^{*}=(\underbrace{V\times \cdots\times V}_{r-\text{copies}})^{*}\neq \underbrace{V^{*}\otimes\cdots\otimes V^{*}}_{r-\text{copies}}=T^{r}$

Long Answer: The previous question/section established that $(V^{r})^{*}\neq T^{r}.$ In writing $T^{r}=\underbrace{V^{*}\otimes\cdots\otimes V^{*}}_{r-\text{copies}}$ the author is saying that any multilinear function defined on $V^{r}$ can be expressed as a linear combination of $r$-fold tensor products of elements from the dual basis for $V^{*}$.

For example, suppose $V$ is a 2-dimensional real vector space with basis $\{e_{1}, e_{2}\}$, and that $\{\alpha^{1}, \alpha^{2}\}$ is the corresponding dual basis for $V^{*}.$ The statement that $T^{2}=V^{*}\otimes V^{*}$ means that any $f\in T^{2}$ can be written as $$f = a\cdot \alpha^{1}\otimes \alpha^{1}+b\cdot\alpha^{1}\otimes \alpha^{2}+c\cdot \alpha^{2}\otimes \alpha^{1}+d\cdot \alpha^{2}\otimes \alpha^{2};$$ i.e., $\{\alpha^{i}\otimes \alpha^{j}\}_{i,j=1}^{2}$ is a basis for $T^{2}.$

Note the presence of the cross terms $\alpha^{1}\otimes \alpha^{2}$ and $\alpha^{2}\otimes \alpha^{1}$ above. This is no accident and is directly related to the presence of the cross terms in the Euclidean inner product example above. It may be a useful exercise to try writing the inner product example above in terms of tensor products.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
Hi Peter ,

These are deep, insightful considerations that truly highlight your attention to detail. I think these are great questions because it is easy to overlook the subtlety regarding these topics.



Short Answer: No, $T^{r}\neq (V^{r})^{*}.$

Long Answer: The intuition to try and link $T^{r}$ with $(V^{r})^{*}$ is reasonable and tempting. Unfortunately, this is not the case.


Counterexample 1: Linear Functions Need Not be Multilinear

Consider $V=\mathbb{R}$, and let $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ be defined by $f(x,y)= 2x +3y.$ This is an example of a linear function on $\mathbb{R}^{2}=\mathbb{R}\times\mathbb{R}$ that is not multilinear.


Counterexample 2a: Multilinear Functions Need Not be Linear

Consider $V=\mathbb{R}^{2},$ and let $f:\mathbb{R}^{2}\times \mathbb{R}^{2}\rightarrow\mathbb{R}$ be defined to be the Euclidean inner product; i.e., $f({\bf x}, {\bf y}) = \langle {\bf x}, {\bf y}\rangle.$ This is an example of a multilinear function that is not linear because, using multilinearity, $$f({\bf x}_{1}+{\bf x}_{2}, {\bf y}_{1}+{\bf y}_{2})=f({\bf x}_{1},{\bf y}_{1})+f({\bf x}_{1},{\bf y}_{2})+f({\bf x}_{2},{\bf y}_{1})+f({\bf x}_{2},{\bf y}_{2}).$$ However, linearity would require $$f({\bf x}_{1}+{\bf x}_{2}, {\bf y}_{1}+{\bf y}_{2})= f({\bf x}_{1}, {\bf y}_{1}) + f({\bf x}_{2},{\bf y}_{2}).$$ The presence of the mixed terms $f({\bf x}_{1}, {\bf y}_{2})$ and $f({\bf x}_{2}, {\bf y}_{1})$ is the essence of the difference between multilinear and linear functions on products of vector spaces.


Counterexample 2b: Multilinear Functions Need Not be Linear

Let $V=\mathbb{R}^{n}$, and define $f:\underbrace{\mathbb{R}^{n}\times\cdots\times\mathbb{R}^{n}}_{n-\text{copies}}$ to be the determinant function; i.e., $f({\bf v}_{1},\ldots, {\bf v}_{n})=\text{Det}[{\bf v}_{1},\ldots, {\bf v}_{n}].$ It is know from the properties of determinants that $f$ is multilinear (see https://en.wikipedia.org/wiki/Determinant), but that $f$ is not linear because, in general, $\text{Det}(A+B)\neq \text{Det}(A)+\text{Det}(B)$.


Standard Multilinear Function Examples

The archetype of multilinear functions are the Euclidean inner product on $\mathbb{R}^{n}\times\mathbb{R}^{n}$, the cross product on $\mathbb{R}^{3}\times\mathbb{R}^{3}$, and the determinant on $\underbrace{\mathbb{R}^{n}\times\cdots\times\mathbb{R}^{n}}_{n-\text{copies}}.$




Short Answer: $(V^{r})^{*}=(\underbrace{V\times \cdots\times V}_{r-\text{copies}})^{*}\neq \underbrace{V^{*}\otimes\cdots\otimes V^{*}}_{r-\text{copies}}=T^{r}$

Long Answer: The previous question/section established that $(V^{r})^{*}\neq T^{r}.$ In writing $T^{r}=\underbrace{V^{*}\otimes\cdots\otimes V^{*}}_{r-\text{copies}}$ the author is saying that any multilinear function defined on $V^{r}$ can be expressed as a linear combination of $r$-fold tensor products of elements from the dual basis for $V^{*}$.

For example, suppose $V$ is a 2-dimensional real vector space with basis $\{e_{1}, e_{2}\}$, and that $\{\alpha^{1}, \alpha^{2}\}$ is the corresponding dual basis for $V^{*}.$ The statement that $T^{2}=V^{*}\otimes V^{*}$ means that any $f\in T^{2}$ can be written as $$f = a\cdot \alpha^{1}\otimes \alpha^{1}+b\cdot\alpha^{1}\otimes \alpha^{2}+c\cdot \alpha^{2}\otimes \alpha^{1}+d\cdot \alpha^{2}\otimes \alpha^{2};$$ i.e., $\{\alpha^{i}\otimes \alpha^{j}\}_{i,j=1}^{2}$ is a basis for $T^{2}.$

Note the presence of the cross terms $\alpha^{1}\otimes \alpha^{2}$ and $\alpha^{2}\otimes \alpha^{1}$ above. This is no accident and is directly related to the presence of the cross terms in the Euclidean inner product example above. It may be a useful exercise to try writing the inner product example above in terms of tensor products.


Thanks for the extensive help GJA, I really appreciate it ...

Still reflecting on what you have said ...

Thanks again ...

Peter