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5

Yes: your conjecture can be proven simply by looking at the dimensions of the vector spaces. In particular, $\otimes_k K$ commutes with direct sums, and $$ \left(\bigoplus_i k \right) \otimes_k K \cong \bigoplus_i K $$ so $\dim_k V = \dim_K (V \otimes_k K)$. (note that it is important that you asked for them to be isomorphic as $K$-vector spaces, as $K ...


4

You can define tensor product by the universal property. Then you just need to check that $U\otimes V\oplus U\otimes W$ satisfies the universal property for tensor product of $U$ and $V\oplus W$. This check is pretty much trivial. This approach is very useful, because it allows you to prove a lot of other similar identities, and not necessarily for finite ...


3

We'll avoid the metric tensor altogether to keep things simple. The demonstration is simpler this way, although not "proper". The polar coordinate system is rather "nice" to deal with, so you don't need the full machinery of tensors in curvilinear coordinates. First of all, note that W is a scalar. The second fact is that you can think of polar coordinates ...


3

A proof analogous to the one for tensor products will establish the following more general result that you can find in Milnor and Stasheff's Characteristic Classes: Let $\mathcal{V}$ be the category of finite dimensional real vector spaces and isomorphisms of such (not all linear transformations, just isomorphisms, for maximum generality). Let $T : ...


2

Here are some of the details implicit in @Sasha's answer. Notation: Given vectors spaces $X,Y$ and elements $x \in X$, $y \in Y$, I'll write $x \oplus y$ for the corresponding element of the vectors space direct sum $X \oplus Y$. The map \begin{align*} U \times (V \oplus W) \to (U \otimes V) \oplus (U \otimes W) && (u, v \oplus w) \mapsto ...


2

Your first map can be written as $$F\left(\sum_{i=1}^n u_i\otimes(v_i,w_i)\right)=\left(\sum_{i=1}^n u_i\otimes v_i,\sum_{i=1}^n u_i\otimes w_i\right)$$ Now the inverse is $$F^{-1}\left(\sum_{i=1}^n u_i\otimes v_i,\sum_{j=1}^m u'_j\otimes w_j\right)=\sum_{i=1}^n u_i\otimes (v_i,0)+\sum_{j=1}^m u'_j\otimes (0,w_j).$$


2

The problem is, the differential operator doesn't commute with functions. It is unreasonable to write $\partial_i\gamma\partial_i\gamma=\partial_i\partial_i\gamma\gamma$. A common way to write this $$\delta^{ij}\partial_i\gamma\partial_j\gamma=\partial^j\gamma\partial_j\gamma$$


1

No, you need the tensor product $\otimes$ to construct the most general tensors. The wedge product $\wedge$ produces skew-symmetric objects. Well, the space of tensors on a vector space V is constructed by taking tensor products of the form $V \otimes V \otimes \ldots V \otimes V^* \otimes V^* \ldots V^*$ where $V$ is the vector space and $V^*$ is its ...


1

Of course, no. All non-degenerate symmentic bilinear forms over $\mathbb{C}$ are the same. But not over $\mathbb{R}$.


1

Hint: pick bases for $U,V,W$ to induce bases for $U\otimes(V\oplus W)$ and $(U\otimes V)\oplus (U\otimes W)$. How does this prove the obvious map is an isomorphism between them?


1

It is enough to describe $\mathbb{R} \otimes_{\mathbb{Z}} \mathbb{R}$. If $c=|\mathbb{R}|$, there is an isomorphism of $\mathbb{Q}$-vector spaces, hence of abelian groups $\mathbb{R} \cong \mathbb{Q}^{(c)}$. It follows $\mathbb{R} \otimes_{\mathbb{Z}} \mathbb{R} \cong \mathbb{Q}^{(c \times c)} \cong \mathbb{Q}^{(c)} \cong \mathbb{R}$. Notice that this is ...


1

A priori, $A \otimes_K L$ does not have an $L$-module structure. However, the equation you have above defines an $L$-module structure on it. In other words, this is a definition, not a proposition.



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