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Why is a dueldual vector a type (1,0) tensor?

I am reading "An introduction to Spacetime and Geometry" by Sean Carroll, and on page 21 section 1.6, it is mentioned that the dual of a vector is a type (0,1) tensor.

The definition given of the page for a tensor is as follows (1):

\begin{equation}\tag{1} T:T^*_p\underbrace{\times...\times}_{k \text{ times}}T^*_p\times T_p\underbrace{\times...\times}_{l \text{ times}}T_p\mapsto\mathbb{R} \end{equation}

In (1), $T_p$ stands for the tangent space at a point of a manifold $M$, and $T^*_p$ the dual to the same space. $\times$ denotes the Cartesian product.

By this definition (1), a $(k,l)$ tensor would have $k$ dueldual vectors and $l$ regular vectors.

However, the textbook claims that a dual vector can be thought of as a type (0,1) tensor and a regular one a type (1,0). This seems to be in contradiction of (1), where dual vectors appear $k$-times and regular ones $l$-times.

Which leads to my question:

Have I misinterperated the definition? If not, why are the indices of dual and regular vectors switched for a type $(k,l)$ tensor?

Why is a duel vector a type (1,0) tensor?

I am reading "An introduction to Spacetime and Geometry" by Sean Carroll, and on page 21 section 1.6, it is mentioned that the dual of a vector is a type (0,1) tensor.

The definition given of the page for a tensor is as follows (1):

\begin{equation}\tag{1} T:T^*_p\underbrace{\times...\times}_{k \text{ times}}T^*_p\times T_p\underbrace{\times...\times}_{l \text{ times}}T_p\mapsto\mathbb{R} \end{equation}

In (1), $T_p$ stands for the tangent space at a point of a manifold $M$, and $T^*_p$ the dual to the same space. $\times$ denotes the Cartesian product.

By this definition (1), a $(k,l)$ tensor would have $k$ duel vectors and $l$ regular vectors.

However, the textbook claims that a dual vector can be thought of as a type (0,1) tensor and a regular one a type (1,0). This seems to be in contradiction of (1), where dual vectors appear $k$-times and regular ones $l$-times.

Which leads to my question:

Have I misinterperated the definition? If not, why are the indices of dual and regular vectors switched for a type $(k,l)$ tensor?

Why is a dual vector a type (1,0) tensor?

I am reading "An introduction to Spacetime and Geometry" by Sean Carroll, and on page 21 section 1.6, it is mentioned that the dual of a vector is a type (0,1) tensor.

The definition given of the page for a tensor is as follows (1):

\begin{equation}\tag{1} T:T^*_p\underbrace{\times...\times}_{k \text{ times}}T^*_p\times T_p\underbrace{\times...\times}_{l \text{ times}}T_p\mapsto\mathbb{R} \end{equation}

In (1), $T_p$ stands for the tangent space at a point of a manifold $M$, and $T^*_p$ the dual to the same space. $\times$ denotes the Cartesian product.

By this definition (1), a $(k,l)$ tensor would have $k$ dual vectors and $l$ regular vectors.

However, the textbook claims that a dual vector can be thought of as a type (0,1) tensor and a regular one a type (1,0). This seems to be in contradiction of (1), where dual vectors appear $k$-times and regular ones $l$-times.

Which leads to my question:

Have I misinterperated the definition? If not, why are the indices of dual and regular vectors switched for a type $(k,l)$ tensor?

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Why is a duel vector a type (1,0) tensor?

I am reading "An introduction to Spacetime and Geometry" by Sean Carroll, and on page 21 section 1.6, it is mentioned that the dual of a vector is a type (0,1) tensor.

The definition given of the page for a tensor is as follows (1):

\begin{equation}\tag{1} T:T^*_p\underbrace{\times...\times}_{k \text{ times}}T^*_p\times T_p\underbrace{\times...\times}_{l \text{ times}}T_p\mapsto\mathbb{R} \end{equation}

In (1), $T_p$ stands for the tangent space at a point of a manifold $M$, and $T^*_p$ the dual to the same space. $\times$ denotes the Cartesian product.

By this definition (1), a $(k,l)$ tensor would have $k$ duel vectors and $l$ regular vectors.

However, the textbook claims that a dual vector can be thought of as a type (0,1) tensor and a regular one a type (1,0). This seems to be in contradiction of (1), where dual vectors appear $k$-times and regular ones $l$-times.

Which leads to my question:

Have I misinterperated the definition? If not, why are the indices of dual and regular vectors switched for a type $(k,l)$ tensor?