# A Basis for T^r, the space of covariant tensors of rank r on V ... ... Browder Theorem 12.9 ...

#### 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 proof of Theorem 12.9 on pages 271-272 ... ...

The relevant text (Theorem 12.8 together with the preceding definition, Definition 12.8) reads as follows:

In the above text from Browder, at the start of the proof of Theorem 12.9, we read the following:

" ... ... It is immediate from the definition of the tensor product that

$$\displaystyle \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} ( u_{k_1} ..... u_{k_r} ) = \delta^{ j_1 \cdot \cdot \cdot j_r }_{ k_1 \cdot \cdot \cdot k_r }$$

... ... ... "

$$\displaystyle \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} ( u_{k_1} ..... u_{k_r} ) = \delta^{ j_1 \cdot \cdot \cdot j_r }_{ k_1 \cdot \cdot \cdot k_r }$$

is true ... ... ?

(Please note that I am slightly lost and overwhelmed by the indices in this theorem ... )

Hope someone can help ...

Peter

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I am uncertain of the nature of the $$\displaystyle \tilde{u}^{j_i}$$ ... but a basis with similar notation was given in the proof of Theorem 12.2 ... so I am providing the text of Chapter 12 up to and including Theorem 12.9 to give readers the context and notation of the Chapter ... ... and any necessary preliminary definitions and results ... as follows:

Hope that helps ...

Peter

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#### steenis

##### Well-known member
MHB Math Helper
Maybe it is an idea to practice first with
$$n=1, r=1$$
then
$$n=2, r=1$$
$$n=2, r=2$$
$$n=3, r=1$$
$$n=3, r=2$$
$$n=7, r=3$$
and so on

#### Peter

##### Well-known member
MHB Site Helper
Maybe it is an idea to practice first with
$$n=1, r=1$$
then
$$n=2, r=1$$
$$n=2, r=2$$
$$n=3, r=1$$
$$n=3, r=2$$
$$n=7, r=3$$
and so on

THanks Steenis ...

Good idea ...

Peter

#### Peter

##### Well-known member
MHB Site Helper
Maybe it is an idea to practice first with
$$n=1, r=1$$
then
$$n=2, r=1$$
$$n=2, r=2$$
$$n=3, r=1$$
$$n=3, r=2$$
$$n=7, r=3$$
and so on

Trying Steenis' idea ...

Practice values for first statement in the proof of Theorem 12.9 in Browder ...

... that is practice values for n and r in both statement (12.1) and

$$\displaystyle \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} ( u_{k_1} ..... u_{k_r} ) = \delta^{ j_1 \cdot \cdot \cdot j_r }_{ k_1 \cdot \cdot \cdot k_r }$$

Now ... ... try $$\displaystyle n = 1, r = 1$$ ...

Then the basis for $$\displaystyle V$$ is $$\displaystyle \{ u_1 \}$$

... ... and the basis for $$\displaystyle T^r$$ is $$\displaystyle \{ \tilde{u}^1 \}$$ ... ...

now consider

$$\displaystyle \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} ( u_{k_1} ..... u_{k_r} ) = \delta^{ j_1 \cdot \cdot \cdot j_r }_{ k_1 \cdot \cdot \cdot k_r }$$ ... ... ... (2)

[NOTE: Problem with (2): It appears to me that $$\displaystyle ( u_{k_1} ..... u_{k_r} )$$ means take any $$\displaystyle r$$ of the $$\displaystyle n$$ basis vectors of $$\displaystyle V$$ ... ... but does the order matter ... for example do we include $$\displaystyle (u_1, u_2)$$ and $$\displaystyle (u_2, u_1)$$ ... or are they counted as the same ... ... note that since it is an ordered list I suspect they are different ... is that right? ... ... ]

From (2) we have $$\displaystyle \tilde{u}^{j_1} ( u_{ k_1} ) = \tilde{u}^1 ( u_1 ) = 1$$

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Try $$\displaystyle n = 2, r = 1$$

Then the basis for $$\displaystyle V$$ is $$\displaystyle ( u_1, u_2 )$$

... ... and the basis for $$\displaystyle T^r$$ is $$\displaystyle \{ \tilde{u}^{ j_1 } \ : \ 1 \leq j_1 \leq 2 \} = \{ \tilde{u}^1, \tilde{u}^2 \}$$ ... ...

then we consider

$$\displaystyle \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} ( u_{k_1} ..... u_{k_r} ) = \delta^{ j_1 \cdot \cdot \cdot j_r }_{ k_1 \cdot \cdot \cdot k_r }$$ ... ... ...

$$\displaystyle = \tilde{u}^{j_1} ( u_{k_1} )$$ where $$\displaystyle k_1 = u_1$$ or $$\displaystyle u_2$$ and $$\displaystyle \tilde{u}^{j_1} ( u_{k_1} ) = \delta^{ j_1 }_{k_1}$$

Explicitly we have

$$\displaystyle \tilde{u}^1 ( u_1 ) = 1$$

$$\displaystyle \tilde{u}^1 ( u_2 ) = 0$$

$$\displaystyle \tilde{u}^2 ( u_1 ) = 0$$

$$\displaystyle \tilde{u}^2 ( u_2 ) = 1$$

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Try $$\displaystyle n = 2, r = 2$$

Then the basis for $$\displaystyle V$$ is $$\displaystyle ( u_1, u_2 )$$

... ... and the basis for $$\displaystyle T^r$$ is $$\displaystyle \{ \tilde{u}^{ j_1 } \otimes \tilde{u}^{ j_2 } \ : \ 1 \leq j_1 \leq 2 1 \leq j_2 \leq 2 \} = \{ \tilde{u}^1, \tilde{u}^2 \}$$ ... ...

[Note : Dimension of $$\displaystyle T^r = n^r = 2^2 = 4$$ ... ... ]

then we consider

$$\displaystyle \tilde{u}^{ j_1 } \otimes \tilde{u}^{ j_2 } ( u_{ k_1 }, u_{ k_2 } )$$

Possible values

$$\displaystyle \tilde{u}^1 \otimes \tilde{u}^1 ( u_1, u_2 ) = \tilde{u}^1 ( u_1 ) \tilde{u}^1 ( u_2 ) = 1.0 = 0$$

$$\displaystyle \tilde{u}^2 \otimes \tilde{u}^1 ( u_1, u_2 ) = \tilde{u}^2 ( u_1 ) \tilde{u}^1 ( u_2 ) = 0.0 = 0$$

$$\displaystyle \tilde{u}^1 \otimes \tilde{u}^2 ( u_1, u_2 ) = \tilde{u}^1 ( u_1 ) \tilde{u}^2 ( u_2 ) = 1.1 = 1$$

$$\displaystyle \tilde{u}^2 \otimes \tilde{u}^2 ( u_1, u_2 ) = \tilde{u}^2 ( u_1 ) \tilde{u}^2 ( u_2 ) = 0.1 = 0$$

[ Problem: Do we have to include $$\displaystyle \tilde{u}^{ j_1 } \otimes \tilde{u}^{ j_2 } ( u_2, u_1 )$$ in the above list

Note that in

$$\displaystyle \tilde{u}^1 \otimes \tilde{u}^2 ( u_1, u_2 ) = \tilde{u}^1 ( u_1 ) \tilde{u}^2 ( u_2 ) = 1.1 = 1$$

we have $$\displaystyle ( j_1, j_2) = ( k_1, k_2) = (1, 2 )$$ ....

Can someone please confirm that the above is correct ... or alternatively point out any errors or shortcomings ...

Peter

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