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

#### Peter

##### Well-known member
MHB Site Helper
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:

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
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
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