# What does “$(n,m)$-tensor” mean?

I know the meaning of tensor, but I forgot the meaning of "$(n,m)$-tensor". What do $n$ and $m$ refer to?

Thanks.

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What do you mean by tensor: an element of a tensor product of vector spaces (if not modules) or a field of tensors on a manifold? Assuming your meaning of tensor is the first one, an $(n,m)$ tensor means an element of $(V^{\otimes n}) \otimes (V^*)^{\otimes m}$ for some vector space $V$. –  KCd Jun 27 '12 at 11:39
An $(n,m)$-tensor on a finite-dimensional real vector space $V$ is (usually) defined to be a multilinear map $\Phi:\underbrace{V^{\ast}\times\cdots \times V^{\ast}}_{n\text{ times}}\times \underbrace{V\times\cdots \times V}_{m\text{ times}}\to \mathbb{R}$; $V^{*}$ denotes the dual space of $V$, i.e., the real vector space of all linear functionals $V\to\mathbb{R}$. The nonnegative integers $n$ and $m$ are referred to as the covariant and contravariant orders of the type $(n,m)$-tensor $\Phi$ on $V$, respectively.
You have the $V$ and $V^*$ backwards. For example, a tensor of type $(n,0)$ is supposed to be an element of $V^{\otimes n}$, but by your description it is a multilinear map $V \times \cdots \times V \rightarrow {\mathbf R}$, which in terms of tensor products is an element of $(V^{\otimes n})^* \cong (V^*)^{\otimes n}$. –  KCd Jun 27 '12 at 11:36
You need $V$ to be finite-dimensional here. A tensor of type $(n, m)$ is supposed to be an element of $V^{\otimes n} \otimes (V^*)^{\otimes m}$. –  Zhen Lin Jun 27 '12 at 13:06
You say this was fixed, but I do not see a fix having been made. A (1,0) tensor is just an element of $V$, but by the way the answer is currently written (which may change after I post this) a (1,0) tensor is a linear map from $V$ to ${\mathbf R}$, which is an element of $V^*$. –  KCd Jun 27 '12 at 15:29
@KCd How about now? The reason I said it was fixed earlier was because I thought you were saying that my definition of $n$ and $m$ as the covariant and contravariant orders of the tensor $\Phi$ on $V$ was incorrect; rather $n$ and $m$ are the contravariant and covariant orders of $\Phi$ on $V$ (and I agree with that). I assumed that otherwise the notation "$(n,m)$" for a type $(n,m)$-tensor $\Phi$ on $V$ was a matter of convention and different people used different conventions to denote the same object. However, perhaps I was incorrect; in any case, I have changed my answer. –  Amitesh Datta Jun 28 '12 at 3:01